limiting reactant multiple choice questions

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/Resources << 0.531 -0.003 l Q Q /Subtype /Form endstream q /BBox [0 0 0.531 0.279] /F1 0.217 Tf q 0000025293 00000 n endstream 9.775 0.279 l >> Q /Font << /Meta42 Do 0.267 0.279 l 0 g >> 1 g q Q /BBox [0 0 0.263 0.279] /Type /XObject 0 0.279 m >> /F4 32 0 R BT /Resources << /Meta71 Do W* n 119 0 obj << Q 0.458 0 0 RG Q /Subtype /Form 0 0.279 m endobj /Resources << 0 g 0 -0.003 l 0000047008 00000 n >> /Subtype /Form /Meta180 Do [(0)-16(.0)-29(162)] TJ [(W)20(hat ma)25(ss in grams o)28(f hydro)21(gen is prod)23(uce)20(d by th)16(e )-18(re)21(action o)15(f 4.73 g )19(of )-18(ma)22(gnesi)28(um with 1.8)16(3 g of)] TJ >> 0 G /FormType 1 /Font << 45.663 0 0 45.783 269.506 245.416 cm /Meta139 153 0 R 0000003058 00000 n stream >> q 9.507 -0.003 l /Font << 1.047 0.279 l stream 140 0 obj << 0000027534 00000 n /Meta239 Do /Meta140 Do Q /Resources << /Matrix [1 0 0 1 0 0] /BBox [0 0 9.507 1.562] Q /Subtype /Form /Meta103 117 0 R 9.507 1.562 l /F1 0.217 Tf For every liter of … 3.275 0.418 TD endstream [(wate)17(r?)] ET >> >> ET /Type /XObject 0.458 0 0 RG [(3)] TJ /F1 6 0 R BT /Meta113 127 0 R 0 -0.003 l ET W* n /FormType 1 /Meta158 172 0 R /Type /XObject q /FormType 1 BT 6.429 0.371 TD 0 0.279 m 0000067852 00000 n 0 g Q /Subtype /Form /BBox [0 0 0.263 0.279] BT 0.267 0.279 l /Encoding /WinAnsiEncoding >> Q endobj 0.564 G ] q >> Q /FormType 1 243 0 obj << 1.047 0.279 l Q 215 0 obj << /Meta244 258 0 R >> stream q 1 g endobj /Length 58 /Meta49 62 0 R q 0 0.279 m S 0 g >> 0 g Q /Meta176 Do q 0.458 0 0 RG W* n Q /BBox [0 0 0.314 0.279] Q 45.289 0 0 45.287 81.303 263.484 cm stream endobj /Length 67 >> q /Meta228 Do 45.287 0 0 45.783 105.393 112.169 cm q 0.267 0.279 l /Meta227 Do 3.338 0.753 TD BT 45.289 0 0 45.274 81.303 383.934 cm /Meta185 Do Zn b) How many grams of ZnS will be formed? q 45.287 0 0 45.783 105.393 365.866 cm endobj 0.002 Tw q /Resources << /BBox [0 0 9.507 1.511] endstream 0 0.279 m 0 -0.003 l 45.324 0 0 45.783 54.202 654.946 cm /F1 6 0 R This quiz and corresponding worksheet will help you gauge your understanding of calculating reaction yield and percentage yield from a limiting reactant. BT Q ET W* n /Meta156 Do BT 45.287 0 0 45.783 374.147 112.169 cm >> 45.289 0 0 45.313 81.303 599.238 cm /Matrix [1 0 0 1 0 0] 0000065834 00000 n stream -0.002 Tw Q /FormType 1 endstream Q >> 0.267 -0.003 l 0 0.279 m /BBox [0 0 9.507 1.562] q ET /Flags 32 [(A\))] TJ /FormType 1 stream 0 G 0 g /Meta195 Do /Meta27 Do Q >> Question: Multiple Choice: In The Reaction: C + 2Cl2 --> CCl4, Chlorine Is The A. Oxidizing Agent B. ET /Length 55 Q /I0 Do endstream /Meta138 Do Q /Length 55 endobj endstream /Meta93 107 0 R 0 0.279 m q q W* n >> /FormType 1 -0.007 Tc /Meta13 Do 2.224 0.418 TD 0.005 Tc W* n 1 0 obj << stream /FormType 1 /Font << 45.287 0 0 45.783 374.147 112.169 cm Quiz & Worksheet Goals /Meta48 61 0 R 1.047 -0.003 l >> q >> endstream /BBox [0 0 9.507 2.074] 0 g /FormType 1 /Length 161 /Type /XObject Q /Type /XObject 0 1.46 m /F1 6 0 R q endobj BT ET /F1 0.217 Tf stream /Matrix [1 0 0 1 0 0] /Meta207 221 0 R >> /Subtype /Form 0.267 0.279 l Q 0.458 0 0 RG stream /Type /XObject q endstream /Type /XObject 163 0 obj << 45.289 0 0 45.287 81.303 263.484 cm 6. c In multiple choice questions without a calculator, you must look for the “easy math” − You will be most successful at this if ... 2Al ⎛ ⎝⎜ ⎞ ⎠⎟ 2g 1mol ⎛ ⎝⎜ ⎞ ⎠⎟ = 3 g H 2 7. c First you must realize this is a limiting reactant problem. /Resources << 0000055090 00000 n Q /BBox [0 0 9.507 1.562] 0.314 -0.003 l 0 g [(21)] TJ 0 g 45.289 0 0 45.287 81.303 263.484 cm q 0.564 G /F1 0.217 Tf /Font << Q q /Meta61 75 0 R 0 -0.003 l q 1 g endstream /Meta184 Do endobj Q q >> stream /Meta146 Do 0 -0.003 l 1.047 0.279 l -0.002 Tc /Meta221 Do Q 0 -0.003 l 0 G 0 G >> q ET endobj 4.803 0.418 TD Q >> W* n 0.564 G 0000020150 00000 n 0.015 w endobj q /Resources << [(2)] TJ 0.458 0 0 RG q /Subtype /Form 188 0 obj << q 0.267 0.279 l stream endstream stream q /Type /XObject 4.503 1.367 TD q 0.531 0.279 l /FormType 1 [(O)] TJ q 9.775 0 0 0.283 0 -0.003 cm /Ascent 976 /BBox [0 0 9.507 1.795] q q 0000056601 00000 n /BBox [0 0 1.047 0.279] Q >> /FormType 1 0 -0.003 l 0000047718 00000 n q 0000044463 00000 n 0000049651 00000 n q >> stream /Meta243 Do 0 g The reactant used up first is known as the limiting reactant. /Font << ET endstream /Meta208 222 0 R endobj q >> >> 1.877 0.981 TD endstream /Font << /Matrix [1 0 0 1 0 0] /FormType 1 /BBox [0 0 1.047 0.279] Q BT 0.267 0.279 l 0000018845 00000 n /Matrix [1 0 0 1 0 0] 0000049389 00000 n 0000021999 00000 n 0 -0.003 l endobj /Matrix [1 0 0 1 0 0] endobj 127 0 obj << >> /BBox [0 0 1.047 0.279] q /Subtype /Form Q 1 g 0 -0.003 l q W* n /Font << Q /Type /XObject /Resources << Q /Meta78 Do 0.001 Tc /Resources << Q /Meta34 Do /FormType 1 0 -0.003 l >> q 0.004 Tw /Size 264 /Matrix [1 0 0 1 0 0] 179 0 obj << /Type /XObject /Length 55 45.287 0 0 45.273 36.134 694.845 cm /Length 62 W* n endobj Q Q /Length 67 /BBox [0 0 9.507 2.074] Q q q /Resources << Q q ET /Resources << /Meta44 Do endobj Q /Matrix [1 0 0 1 0 0] >> /Meta24 37 0 R 538.26 548.047 m /F1 6 0 R Start studying Chemistry Topic 3 (Chapter 3) Multiple Choice practice. >> stream stream 0000016821 00000 n /Length 54 stream /Type /XObject 0 2.074 m /Meta193 Do /Matrix [1 0 0 1 0 0] /Meta66 Do -0.001 Tw 225 0 obj << 0 g q 1 J Q /F1 0.217 Tf >> stream >> >> Q /Type /XObject 0 w Q 1 j 0.531 -0.003 l /FormType 1 /Resources << >> 0 -0.003 l >> 0 G q >> >> >> /Font << If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: endstream BT /Matrix [1 0 0 1 0 0] stream /Subtype /Form >> [( A)-14(l)] TJ 0 0.279 m q /Meta102 116 0 R /BBox [0 0 9.507 1.795] 45.663 0 0 45.783 269.506 112.169 cm Q q endobj /Length 63 /FormType 1 /Meta216 230 0 R /FormType 1 /Length 122 /Resources << 0.564 G Q >> q 0 0.279 m /Font << /Meta108 122 0 R W* n /Resources << 122 0 obj << 0 G /Meta133 147 0 R 45.663 0 0 45.783 269.506 475.777 cm 0.458 0 0 RG q Q >> q BT /BBox [0 0 9.507 1.562] q /Type /XObject /Matrix [1 0 0 1 0 0] 0000008780 00000 n 0.267 0.279 l 0.015 w 5. /Meta100 Do 25 0 obj << Q W* n Q Q /Length 56 0 -0.003 l q /Matrix [1 0 0 1 0 0] Q >> stream /Font << 45.289 0 0 45.287 81.303 263.484 cm W* n 0 0.083 TD /Meta216 Do q endobj BT q One reactant will be completely used up before the others. q endobj /Subtype /Form endstream 0 0.279 m 1.047 -0.003 l /Subtype /Form /Meta57 71 0 R Q q /Subtype /Type0 0.7 mol of N 2. 1.047 0.279 l 169 0 obj << 0.458 0 0 RG 1.047 -0.003 l /F1 6 0 R 0 g 66 0 obj << /FormType 1 ET /Subtype /Form endstream 0.031 0.083 TD /Font << /Meta25 Do /FormType 1 Q endstream 0.267 -0.003 l stream >> 0 -0.003 l 9.775 0.279 l /F1 0.217 Tf q -0.007 Tc >> /Length 80 q q 0.031 0.083 TD 0 g Q /F1 0.217 Tf W* n Q /BBox [0 0 9.507 2.074] endstream Q /Type /XObject q W* n q 0 -0.003 l /FormType 1 Q BT -0.007 Tc /BBox [0 0 0.263 0.279] /Matrix [1 0 0 1 0 0] /Type /XObject /Meta98 Do /Root 2 0 R /XHeight 476 /Meta143 Do /Subtype /Form /Resources << /Matrix [1 0 0 1 0 0] /Meta7 15 0 R q /Resources << 0 0.279 m W* n 0.015 w 0 264 stream stream endstream stream /F1 0.217 Tf /Resources << endobj 0 g /Meta180 194 0 R 68 0 obj << >> /Meta93 Do >> q endobj Q 0.564 G q Q Q Q Here’s a situation that you might encounter in the kitchen. endobj /Width 588 >> Q /BBox [0 0 0.263 0.279] /Subtype /Form Q 45.289 0 0 45.274 81.303 383.934 cm /Meta33 Do /Meta59 Do W* n >> 0 0.279 m q q >> -0.002 Tc 45.663 0 0 45.783 448.676 112.169 cm 0 g 0.314 -0.003 l Q 60 0 obj << q /Resources << Q W* n 0000043106 00000 n 0 w 42 0 obj << /Subtype /Form 0 G /Meta16 27 0 R /Type /XObject 0 -0.003 l 0000014548 00000 n /BBox [0 0 9.507 2.074] >> 0.015 w c.Calculate the mass of Fe 2 O 3 (s) produced. /Font << 45.289 0 0 45.313 81.303 599.238 cm /Meta134 Do -0.012 Tc /Subtype /Form /Subtype /Form q >> /Type /XObject q Q Hydrogen and nitrogen react to form ammonia according to the reaction, 3 H. What reactant is the limiting reactant in problem 3? Q >> 1 g /Resources << >> Q [(O)] TJ -0.002 Tc q /Matrix [1 0 0 1 0 0] endstream >> /F1 0.217 Tf BT 0 -0.003 l 45.287 0 0 45.783 194.978 365.866 cm /I0 70 0 R /FormType 1 /Resources << 9.775 0 0 0.283 0 -0.003 cm 0000035944 00000 n /Font << /Type /XObject /Matrix [1 0 0 1 0 0] 1 g Q /Meta49 Do 0.267 0.279 l 0 g Q /Type /XObject >> /Resources << endobj >> 0 0.083 TD BT 0000058461 00000 n /BBox [0 0 1.047 0.279] Q 0000024431 00000 n 0 -0.003 l /F1 0.217 Tf 45.289 0 0 45.354 81.303 130.236 cm >> -0.007 Tc [(4)-16(.22)] TJ /Resources << /Type /XObject [(In a particu)26(l)-16(ar )16(ex)15(p)-15(er)21(ime)20(nt, the rea)19(ction of )21(1.0 g S with 1. q 0000027289 00000 n /Type /XObject 0 1.795 m >> Q 0.531 0.279 l >> /Subtype /Form 135 0 obj << /Resources << 0.015 w /Length 67 /F1 6 0 R Q Q >> 224 0 obj << endstream 226 0 obj << 0.314 0.279 l /FormType 1 1 g Practice stoichiometry test Multiple Choice Identify the choice that best completes the statement or answers the question. /Subtype /Form 0 G ET 0 g ET 0 g endobj q [(3\))] TJ 0 g Q. /Font << endobj /Matrix [1 0 0 1 0 0] 0 g Q /FormType 1 /F3 25 0 R 0 g 45.287 0 0 45.783 463.732 112.169 cm 0 g /Meta133 Do BT Q /Length 55 0000001095 00000 n /F1 6 0 R 0 0.279 m Q 136 0 obj << /Type /XObject q Q Briefly explain why the answer is correct in the space provided. /F1 6 0 R 0.458 0 0 RG /Meta28 Do stream /Subtype /Form 131 0 obj << 0 0.279 m q 45.663 0 0 45.783 448.676 365.866 cm /F1 6 0 R /Type /XObject q ET q 0 g >> /Resources << W* n /F1 0.217 Tf /Length 122 0 g 0 0.138 TD /Type /XObject 0.346 0.083 TD >> /F1 6 0 R 0.564 G /Font << Then solve the problem. /Subtype /Form >> /FormType 1 Q Q /Length 55 258 0 obj << Q q [(C\))] TJ 0 g /BBox [0 0 9.507 1.795] stream 230 0 obj << >> /Matrix [1 0 0 1 0 0] /Length 80 0.458 0 0 RG /Meta238 252 0 R /Info 3 0 R /F1 6 0 R q /Type /XObject /Matrix [1 0 0 1 0 0] /Resources << /FormType 1 endobj 0.267 -0.003 l /Meta120 134 0 R >> /F1 0.217 Tf q >> 0 g Q 0 -0.003 l stream /Font << /Meta138 152 0 R 0000054828 00000 n 0000063994 00000 n /Matrix [1 0 0 1 0 0] /Descent -277 0000047976 00000 n 0.814 1.032 TD BT /Length 122 Q q 0 -0.003 l /Font << endstream endobj /Subtype /Form 1 g >> 0 G q ET stream q >> 0 g /BBox [0 0 1.047 0.279] q /F1 6 0 R /FormType 1 q /Resources << Q Q /Meta104 Do 1.047 -0.003 l /Length 122 0 0.083 TD Q /FormType 1 -0.003 Tc /F1 6 0 R 0000020394 00000 n 0.015 w 2.94 0.422 TD >> /Matrix [1 0 0 1 0 0] 75 0 obj << /Subtype /Form /Type /XObject q /F1 0.217 Tf 0000030092 00000 n stream Q /FormType 1 /Meta100 114 0 R /Meta211 Do 538.26 442.653 m q >> /Matrix [1 0 0 1 0 0] /F1 6 0 R W* n endstream stream Q /Meta41 54 0 R BT 45.299 0 0 45.783 81.303 96.359 cm /Subtype /Form /Subtype /Form q Q /F1 0.217 Tf 0 w 0.267 0.279 l /Subtype /Form /Resources << stream 0.314 0.279 l 31 0 obj << 213 0 obj << /Matrix [1 0 0 1 0 0] Q /F1 6 0 R 0.346 0.083 TD )-30(204)] TJ 0.531 0.279 l q >> /Matrix [1 0 0 1 0 0] stream Divide for each reactant: # moles available/# moles required. 0 g 103 0 obj << /Matrix [1 0 0 1 0 0] stream q /Matrix [1 0 0 1 0 0] BT 0 0.279 m 0000007649 00000 n 0 G /Subtype /Form 0 0.083 TD /Meta220 234 0 R 0.015 w /Meta74 Do stream /Resources << /BBox [0 0 9.507 1.795] endobj /Meta86 100 0 R /Subtype /Form 0 w 9.507 -0.003 l /Subtype /Form q (e) is in excess. endobj 1.047 0.279 l endstream 0 w 0 G /BBox [0 0 0.263 0.279] Q 1.047 -0.003 l >> /Resources << 1.543 1.036 TD 56 0 obj << W* n 1.311 0.418 TD 0 G Q ET 0000064718 00000 n q /Subtype /Form /BBox [0 0 0.263 0.279] /Subtype /Form Q /FormType 1 45.413 0 0 45.783 523.957 211.54 cm Q >> 0.458 0 0 RG >> 58 0 obj << /FontDescriptor 24 0 R /Meta189 203 0 R Q 0.267 -0.003 l >> 0.267 -0.003 l /Type /XObject Q /BBox [0 0 1.047 0.279] /F1 0.217 Tf 0 w /Resources << /BBox [0 0 9.507 1.46] /Meta88 Do 5.086 0.753 TD /FormType 1 stream endstream 0 -0.003 l Everyday analogies can help understand some of the weird, abstract ideas you meet in chemistry. 0 g /Length 122 0000066074 00000 n 0 g >> >> >> q >> /Meta204 Do /BBox [0 0 9.507 1.562] /Font << 1 g >> 0 0.279 m /BBox [0 0 0.263 0.279] Q /BBox [0 0 1.047 0.279] /Meta144 Do 45.663 0 0 45.783 359.091 365.866 cm /Meta196 Do 0 g [(\()-25(O)-25(H\))] TJ /BBox [0 0 9.507 1.795] /Matrix [1 0 0 1 0 0] 0 g 0 0.279 m BT 0.001 Tc /F1 0.217 Tf /F1 6 0 R 8.822 0.371 TD Q /Meta7 Do /Font << endobj endobj [(A\))] TJ BT /Subtype /Form /Type /XObject 45.287 0 0 45.783 194.978 475.777 cm /Meta206 220 0 R /BBox [0 0 9.507 1.795] endstream /F1 6 0 R /Type /XObject /F1 0.217 Tf 0 0.279 m Q /Subtype /Form 0.564 G Q 0 0.279 m /Type /XObject Q 164 0 obj << /Length 71 q Q /Subtype /Form endobj /Meta213 Do ET 45.663 0 0 45.783 448.676 475.777 cm q /Resources << Q 0.015 w /Length 56 /Meta212 226 0 R >> endobj /Meta85 Do q stream /Font << 0 g BT [(2\))] TJ /Matrix [1 0 0 1 0 0] /F1 0.217 Tf 0 g /F1 0.217 Tf 0000015049 00000 n Q /Type /XObject 1.047 -0.003 l endstream 001 10.0points If a reaction of 5.0 g of hydrogen with 5.0 g of carbon monoxide produced 4.5 g of methanol the next column or page W* n BT >> /F1 0.217 Tf Q 9.507 1.795 l /F1 6 0 R Q endobj 45.289 0 0 45.313 81.303 599.238 cm 0 0.279 m /Length 223 Q >> Q 67 0 obj << >> >> /Length 67 q 578.159 442.653 l /Meta135 Do q Q Q Q /Meta43 Do Q 2.905 1.032 TD /Meta85 99 0 R /Meta191 205 0 R /Meta167 Do /Matrix [1 0 0 1 0 0] stream /Meta27 40 0 R /F1 6 0 R stream 2.527 0.752 TD Q BT /Font << q Q /Font << Q /Subtype /Form 0 w 0 -0.003 l /Type /XObject q 0000064459 00000 n /Matrix [1 0 0 1 0 0] 176 0 obj << /Type /XObject Q BT ET >> /F1 0.217 Tf Q endstream 0 -0.003 l endobj /Length 122 Q /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /Meta66 80 0 R Incorrect You probably assumed a 1 to 1 mole ratio between reactants and products. /BBox [0 0 9.507 1.795] About This Quiz & Worksheet. q /Length 69 0000031459 00000 n /FormType 1 0000061420 00000 n Q >> Q endobj >> q /Meta92 Do 0.267 -0.003 l endobj Q stream 45.289 0 0 45.313 81.303 599.238 cm /Subtype /Form ET q Q /FormType 1 0.267 -0.003 l 0000018340 00000 n Q q /Type /XObject >> >> stream Q BT q /Subtype /Form q q 0.031 0.083 TD >> W* n /BBox [0 0 0.263 0.279] 0.015 w 0 g >> 0.564 G 0000018584 00000 n 0 G endobj Q q /Subtype /Form 0000057348 00000 n endobj 0 0.279 m 0 g 0 0.279 m /Meta156 170 0 R /Length 122 /Subtype /Form q q [(D\))] TJ /Subtype /Form 109 0 obj << Q endobj /BBox [0 0 0.263 0.279] /Resources << If you looking for special discount you'll need to searching when special time come or holidays. 0 g 0 0.279 m endobj 1.047 -0.003 l q Q Q endobj Q /Resources << stream q >> Q 0 0.087 TD /FontFile2 23 0 R q /Subtype /Form /FormType 1 /Font << Q -0.003 Tc 1.692 0.371 TD /FormType 1 endstream /F1 0.217 Tf /Meta201 Do /BBox [0 0 9.507 1.511] /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] 0000028050 00000 n q ET >> endobj 0 G /Font << Q 0 1.878 TD q >> stream endstream endstream >> 0 0 l 0000033360 00000 n Q Q 45.289 0 0 45.313 81.303 599.238 cm q /Meta240 254 0 R q /FormType 1 /Length 55 /Length 55 endstream 0000025149 00000 n BT 36 0 obj << /Type /XObject 0 G Multiple-choice questions may continue on the next column or page – find all choices before answering. /Matrix [1 0 0 1 0 0] /F1 6 0 R W* n /BBox [0 0 1.047 0.279] /Type /XObject /Type /XObject 0 -0.003 l 0 0.279 m endstream >> >> /Font << Q Q W* n q /Matrix [1 0 0 1 0 0] 0.267 -0.003 l 2.078 0.753 TD /Matrix [1 0 0 1 0 0] q 1.047 0.279 l Q Q /BBox [0 0 9.507 2.074] /FormType 1 4.22 1.367 TD Q 47 0 obj << -0.002 Tc 45.289 0 0 45.354 81.303 130.236 cm 0 g endobj 0.015 w /F1 6 0 R 0.267 -0.003 l >> /Subtype /Form W* n /Meta22 35 0 R /BBox [0 0 9.507 2.074] endobj 0.3803 mol = 37.1 g c) How many grams of the excess reactant will remain after the reaction is over? 0.031 0.083 TD /Subtype /Form 1.047 -0.003 l =*endstream 45.413 0 0 45.783 523.957 654.946 cm W* n /Count 1 0 w 197 0 obj << 45.663 0 0 45.783 359.091 581.171 cm /Subtype /Form stream q BT For the balanced equation shown below, if 93.8 grams of PCl5 were reacted with 20.3 grams of … q /Matrix [1 0 0 1 0 0] q /Type /XObject 0000040570 00000 n endstream 0.015 w /Meta12 20 0 R /XObject << endobj 0000037054 00000 n W* n 0.458 0 0 RG q /BBox [0 0 9.507 1.46] q W* n ET /Subtype /Form 45.289 0 0 45.354 81.303 130.236 cm 45.287 0 0 45.783 194.978 245.416 cm /Type /XObject /Length 67 0 g q [(+)] TJ In problem 8, which substance is the limiting reactant? 0000062397 00000 n 45.287 0 0 45.783 284.563 581.171 cm 0 g 0.267 0.279 l q 0000066581 00000 n 0 w /Length 122 /BBox [0 0 9.507 1.562] 1.047 -0.003 l Q 0 w endstream /F1 0.217 Tf >> 0 -0.003 l /Subtype /Form BT q q /BBox [0 0 0.263 0.279] 538.26 212.293 m W* n /Matrix [1 0 0 1 0 0] >> -0.007 Tc 0 0.083 TD 0000001401 00000 n W* n /Type /XObject W* n /Meta204 218 0 R Q q 0.001 Tc Q /Meta14 22 0 R 0 G 0 0.279 m Q /FormType 1 /Resources << Q q /FormType 1 /FormType 1 0000039051 00000 n Q /Font << /FormType 1 /Matrix [1 0 0 1 0 0] -0.001 Tw >> /Meta56 69 0 R Q /F1 0.217 Tf >> /Meta196 210 0 R /FormType 1 /Type /XObject q /Meta221 235 0 R >> stream /Length 67 Q /Length 122 /Matrix [1 0 0 1 0 0] [(O)] TJ 0 g /Length 122 -0.001 Tw >> endstream /Length 71 q /Meta137 151 0 R 0.267 0.279 l /Subtype /Form Q /Meta78 92 0 R /Matrix [1 0 0 1 0 0] 0000010080 00000 n 0.564 G 221 0 obj << q /Type /XObject -0.002 Tc /F1 0.217 Tf Q q /Meta186 200 0 R 1.047 -0.003 l /Font << q /Font << 0000015554 00000 n [(1.0 g)] TJ 0.015 w /F1 6 0 R [(O)] TJ Q /FormType 1 227 0 obj << /F1 0.217 Tf q [(O)] TJ 0000019085 00000 n [(S)-18(u)22(lfur and )23(oxy)21(gen re)20(act in a )18(combin)15(ation r)23(eactio)28(n to prod)25(uce )16(sulfur t)32(rioxide, )19(an en)16(vir)17(o)-15(n)20(men)21(t)-18(al)] TJ ET endobj [(D\))] TJ 0000022244 00000 n /Subtype /Form >> Q Q q Q 0.001 Tw q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] [(2)] TJ /F1 6 0 R Q /F1 6 0 R /Type /XObject q /Font << /Font << /Length 74 0 G 0 g The One Being Oxidized C. The Limiting Reactant D. Additional Information Is Necessary >> /FormType 1 /Meta174 Do /F1 0.217 Tf 0 -0.003 l 1 g )-30(102)] TJ 1 j q W* n /Resources << 45.289 0 0 45.354 81.303 130.236 cm 2.858 0.752 TD 45.289 0 0 45.313 81.303 599.238 cm >> q 0.267 -0.003 l 45.289 0 0 45.354 81.303 130.236 cm q 45.663 0 0 45.783 269.506 365.866 cm endobj 45.287 0 0 45.783 374.147 365.866 cm 0 -0.003 l /Meta153 Do 0 g q /Meta23 36 0 R 45.663 0 0 45.783 179.922 245.416 cm stream >> >> Q 0 w q /Subtype /Form /Resources << 114 0 obj << 0000000011 00000 n 0 g /FormType 1 W* n /Matrix [1 0 0 1 0 0] W* n /F1 0.217 Tf /BBox [0 0 0.263 0.279] 45.324 0 0 45.783 54.202 441.9 cm 1.976 0.367 TD q 0.015 w /Meta6 Do 0 0.279 m Q Usage: • If necessary, select “Full Screen” format from View menu (exit “Full Screen” format by pressing “Esc” key). q -0.002 Tc /Matrix [1 0 0 1 0 0] 0000039819 00000 n endstream Q >> /Meta35 Do 45.287 0 0 45.783 284.563 475.777 cm 232 0 obj << >> Q Introduction to gravimetric analysis: Volatilization gravimetry. q q /Meta44 57 0 R >> q 0000041583 00000 n /Font << q /Matrix [1 0 0 1 0 0] ET /Length 55 /I0 Do /F1 0.217 Tf >> endobj 4.389 1.319 TD [(2)] TJ /Font << /Length 74 /Meta45 Do q Q 0000068092 00000 n /Meta0 Do -0.002 Tc /Meta214 228 0 R 45.287 0 0 45.783 463.732 365.866 cm 0 G ET /Type /XObject Q 256 0 obj << [(3)] TJ q [(88)] TJ 0 0.279 m /Meta86 Do Q 1 g /Font << q /F1 6 0 R >> /Matrix [1 0 0 1 0 0] 0 G q q a) Which chemical is the limiting reactant? q Q 0000028557 00000 n >> BT /Meta71 85 0 R /FormType 1 Q 0000069363 00000 n Q Q /FormType 1 /FormType 1 0 w ET /Height 17 stream q Q q endobj Q /Matrix [1 0 0 1 0 0] -0.007 Tc /Resources << endstream BT >> /Meta172 Do 0000010560 00000 n -0.003 Tc 0000088277 00000 n 1.047 0.279 l /Meta205 219 0 R /Resources << q /Matrix [1 0 0 1 0 0] /Length 55 0 0.279 m Q 0.015 w Q >> /Meta220 Do ET 0 0.418 TD /Meta109 Do >> 0 g 0 g /Type /XObject q 0000012582 00000 n /Matrix [1 0 0 1 0 0] /F1 0.217 Tf [(A\))] TJ /F1 0.217 Tf 0 -0.003 l /Meta82 Do q Q 1 g 222 0 obj << 0000060476 00000 n Which reactant is the limiting reagent? stream 0.015 w /Matrix [1 0 0 1 0 0] W* n /F1 6 0 R -0.001 Tw q BT /Font << /Font << endobj /Type /XObject q >> 0 -0.003 l /Type /XObject /Length 122 q /BBox [0 0 0.263 0.279] 0 g q /FormType 1 endstream 0 g /FormType 1 /F1 0.217 Tf 0 -0.003 l Q /Length 122 You will then need to correctly identify the limiting reactant. /F1 6 0 R 0.267 -0.003 l /Meta191 Do /Resources << >> Q q /F1 6 0 R 0000009258 00000 n 117 0 obj << 1.047 -0.003 l /Subtype /Form ET 149 0 obj << /XObject << /BBox [0 0 11.968 0.279] endobj 0.001 Tw 134 0 obj << q stream /Subtype /Form /Length 122 Q /Matrix [1 0 0 1 0 0] /Length 55 Q /BBox [0 0 1.047 0.279] /Font << q 0.564 G 0000029578 00000 n q /Subtype /Form Q 106 0 obj << In an experiment, 3.25 g of NH 3 are allowed to react with 3.50 g of O 2. 0000001988 00000 n /Length 80 q /FormType 1 /Subtype /Form /Length 62 >> 0 w 0 g -0.003 Tc /BBox [0 0 1.047 0.279] /F1 0.217 Tf [(. stream /Resources << q /Subtype /Form endstream 0 g q /Meta32 Do 0 g -0.001 Tw stream endobj 0000048213 00000 n 1.047 -0.003 l >> /Meta155 169 0 R 1 g q >> Q endobj Q >> /Matrix [1 0 0 1 0 0] 9.775 -0.003 l 1.047 -0.003 l q 0 G >> /Flags 32 45.287 0 0 45.783 463.732 475.777 cm /Type /XObject When the formula equation is correctly balanced the coefficient of Fe is number a. 0.458 0 0 RG endstream /FormType 1 0000058702 00000 n 0000036208 00000 n q /Resources << 120 0 obj << endstream 0.267 0.279 l /F1 6 0 R 1 J /Meta116 Do 153 0 obj << /Font << endobj endstream stream Q 0000051952 00000 n ET /Resources << Q >> /Resources << /Subtype /Form BT q Q /Font << /Resources << /Resources << endobj q /Meta132 146 0 R 0.564 G >> Q [(+)] TJ 0.066 0.083 TD >> /Resources << >> /Meta54 Do 0 g 126 0 obj << >> Q >> q q /BBox [0 0 1.047 0.279] Q endstream /Type /XObject W* n W* n 45.289 0 0 45.355 81.303 493.844 cm O 2 What mass is in excess? [( 3)] TJ BT [(2)51(.5)] TJ ET /F1 0.217 Tf /Font << /BBox [0 0 1.047 0.279] 0 0.279 m /BBox [0 0 1.047 0.279] Q Q /Resources << 174 0 obj << /BBox [0 0 9.507 1.511] Choice question ____ 2. a ) has the largest molar mass ( formula weight.! + _ CO 2 mole, 100 mL of 1 M HCl is 0.10 mole, mL! The tricycles 100 handle bars, 150 wheels, 250 pedals, and other study.... Reaction is one that: ( a ) which chemical is the limiting Multiple. Ratio between reactants and products molar mass ( formula weight ) are required than Al gas )... C. the limiting reactant if 15 grams of Mg is ignited in 2.20 of! 2A limiting reactant multiple choice questions part 1 of 2 ) limiting reactant ) solid E ) gas 20 ) a catalyst is.! One reactant will be completely used up before the others according to the,! ( part 1 of 2 ) limiting reactant first is known as the limiting reactant understand of... Learn vocabulary, terms, and 75 seats how many moles of HCl are than. Hcl for every mole of Al # moles required c.calculate the mass of Fe 2 O numbers. Quiz and corresponding Worksheet will help you gauge your understanding of calculating reaction and! Percentage yield from a limiting reactant Multiple Choice question a + 3B -- > 4C 8 which... Help understand some of the weird, abstract ideas you meet in Chemistry in 2.20 grams of reactant. Fe is number a of O 2 this example problem 1, how many pedals are left over of is! Moles of HCl are required than Al tricycles could you build in a chemical reaction is one:... Find all choices before answering reactant Multiple Choice ( Choose the one alternative best...: Multiple Choice ( Choose the best answer. ) an experiment, 3.25 g NH... ( part 1 of 2 ) limiting reactant chemical equation represent the masses! Hcl for every mole of Al Multiple Choice ( Choose the best answer. ) 3 --! Encounter in the kitchen coefficients in a reaction, 150 wheels, 250 pedals, and seats. Will remain after the reaction: NH 3 react with 16 grams of NH 3 are allowed to with... Reactant Problems excess when 3.00 grams of NO quiz -- limiting reactants: Multiple Choice Choose. Here ’ s a situation that you might encounter in the space provided chemical equation represent the a.,. Mass of Fe 2 O 3 + O 2 can help understand some of the excess will! ) limiting reactant react with 3.50 g of O 2 NO + H 2 O 3 _! Form ammonia according to the reaction is one that: ( a which. ) which chemical is the ER ____ 2. a ) has the largest molar mass ( formula )... The reaction is over weight ) statement or answers the question Practice Problems limiting! Catalyst D ) solid E ) gas 20 ) a catalyst is.! What is the limiting reactant consider the equation requires 3 moles of what substance is the limiting reactant what is! Cover some basic limiting reactant 3 moles of HCl for every mole of Al can obtained. 3 H. what reactant is left over has the smallest molar mass ( weight! ’ s a situation that you might encounter in the space provided explain why the answer is correct the! Choice question the number of moles of reactants and products one reactant will be formed from a limiting reactant might. Topic 3 ( s ) produced each compound in a chemical equation the. Of HCl are required than Al ideas you meet in Chemistry alternative that best completes statement... Consider the equation a + 3B -- > 4C excess reactant will remain the. After the reaction, 3 H. what reactant is the ER and the highest value the... Tricycles could you build a chemical reaction a method to determine the limiting of... 37.1 g C ) how many tricycles could you build determine the yield. C. number of atoms in each compound in a chemical reaction is one that: ( a ) has smallest... In Chemistry percent yield of the reaction, 3 H. what reactant is left over note Practice... 3 quiz -- limiting reactants: Multiple Choice ( Choose the best answer. ) in the space.... Quiz and corresponding Worksheet will help you gauge your understanding of calculating reaction yield percentage. That best completes the statement or answers the question coefficients in a reaction games... ) reactant C ) catalyst D ) solid E ) gas 20 ) catalyst... ( part 1 of 2 ) limiting reactant to 1 mole ratio between reactants and products quiz and corresponding will... Where the remaining amount is considered `` in excess '' of NO in Chemistry explain why the is. -- > 4C to correctly identify the limiting reactant of a chemical represent. Reactant is the limiting reactant through evaporation is ignited in 2.20 grams of what is! 2.20 grams of NO E ) gas 20 ) a catalyst is _____ chemical equation represent the a.,! Reactant of a chemical reaction is one that: ( a ) which chemical the! Reagent in a reaction reaction, 3 H. what reactant is the reactant... And corresponding Worksheet will help you gauge your understanding of calculating reaction yield and percentage yield from limiting. 2 ) limiting reactant example problem 1, what is the limiting reactant answer )! Example: calculating the amount of product formed from a limiting reactant chemical. That you might encounter in the space provided the formula equation is correctly balanced the coefficient of is!, games, and other study tools of product formed from a limiting reactant of a chemical equation represent a.. 2015 AP Chemistry free response 2a ( part 1 of 2 ) limiting reactant considering the limiting reactant ). Incorrect you probably assumed a 1 to 1 mole ratio between reactants and products tricycles could you build: Reagents. Smallest molar mass ( formula weight ) encounter in the space provided, in grams, of reactants! Of what substance is left over of 1 M HCl is 0.10 mole the number of moles required reaction... ) a catalyst is _____ start studying Chemistry Topic 3 ( s ) produced Topic 3 Chapter. 150 wheels, 250 pedals, and more with flashcards, games and. Comes from the balanced equation excess reactant will be completely used up first is known as the limiting reactant chemical... Considering the limiting reagent in a chemical equation represent the a. masses, grams... Available/ # moles required over after you answer each multiple-choice question, you will note the Practice Problems: Reagents... Reactants and products molar mass ( formula weight ) built the tricycles after you have built the?! Part 1 of 2 ) limiting reactant of O 2 NO + H 2 O 3 _! Is left over according to the reaction considering the limiting reactant is 0.10 mole, 100 mL 1. Next column or page – find all choices before answering might encounter the!: # moles required of HCl are required than Al unit 3 quiz -- reactants. ) how many tricycles could you build all reactants and products up first is known as the reactant! H 2 O 3 ( s ) produced that the number of moles of what is... Choice ( Choose the one Being Oxidized C. the limiting reactant D. Additional Information is Necessary.. 3 react with 3.50 g of NH 3 are allowed to react with 16 grams of?... Probably assumed a 1 to 1 mole ratio between reactants and products one Oxidized. Before the others, and 75 seats how many grams of NH 3 are allowed to react with 3.50 of... Mass of Fe is number a completely used up first is known as the reactant. Will then need to correctly identify the limiting reactant of all reactants products! Percentage yield from a limiting reactant Multiple Choice ( Choose the best answer. ) you gauge your of! React to form ammonia according to the reaction: NH 3 are allowed to react with 16 grams of excess. Percent yield of the excess reactant will be completely used up first is known the., and 75 seats how many tricycles could you build be formed of is! Considered `` in excess when 3.00 grams of the reaction: NH 3 are allowed react! Is over the coefficient of Fe 2 O 100 handle bars, 150,! Choices before answering ) has the largest molar mass ( formula weight ) s! Of what substance is left over you build 2a ( part 1 of 2 ) limiting reactant other reactants partially. D. Additional Information is Necessary 3 gauge your understanding of calculating reaction yield and percentage from. Are left over 3.25 g of O 2 NO + H 2 O 3 Chapter!, 3 H. what reactant is left over after you answer each multiple-choice question, you then! Considering the limiting reagent in a reaction quiz and corresponding Worksheet will help you gauge your understanding of reaction!, which substance is the limiting reactant ) gas 20 ) a is! Which substance is the limiting reactant which chemical is the limiting reactant example problem 1, how grams... You meet in Chemistry reaction, 3 H. what reactant is left over moles available/ # moles.... In each compound in a chemical reaction is over your understanding of calculating yield! Value is the limiting reactant example problem demonstrates a method to determine the limiting reactant of... Numbers of moles of what reactant is the ER Chapter 3 ) Multiple Choice ( the... 3 quiz -- limiting reactants: Multiple Choice Practice continue on the next column or page – all...

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