/BBox [0 0 1.047 0.279] /Length 55 0 0.279 m /Type /XObject /Meta184 198 0 R /BBox [0 0 9.507 1.511] >> q stream /BBox [0 0 1.047 0.279] >> /Meta1 Do /FormType 1 /Length 55 Q [( 2)] TJ /Meta214 228 0 R 0.267 -0.003 l Q q /F1 6 0 R q /F4 0.217 Tf 0 w /FormType 1 0 g 6.165 0.138 TD >> [( S)] TJ 1.047 -0.003 l /Type /XObject -0.001 Tw q 0 0.279 m W* n /F1 6 0 R Q /FormType 1 * What is the limiting reactant if 15 GRAMS of NH 3 react with 16 GRAMS of NO? Q endstream 205 0 obj << /Meta3 9 0 R /Font << /Meta73 Do q 67 0 obj << atoms/molecules/formula units) Chemical formulas or names: Formulas only Names only Mix & match formulas & names Display quiz as: /F1 6 0 R stream /Font << q >> 0.015 w 11.968 -0.003 l 0 -0.003 l stream W* n /Subtype /Image /Subtype /Form Q endobj Q 0 w /Matrix [1 0 0 1 0 0] Q /BBox [0 0 9.507 2.074] q 0 G Q q 45.287 0 0 45.783 463.732 365.866 cm Evaporation. /Type /XObject /Length 161 endstream Q /Matrix [1 0 0 1 0 0] >> >> Q /Meta162 176 0 R How many moles of what reactant is left over? Q Q endstream /Meta94 Do 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 /Meta130 144 0 R 0000019085 00000 n stream /Resources << /Meta74 Do 0000014304 00000 n q q BT /FormType 1 45.289 0 0 45.313 81.303 599.238 cm stream /F1 6 0 R /Meta166 180 0 R endobj /Resources << Q /Subtype /Form 0 g >> endobj Q 1.976 0.367 TD Q 45.299 0 0 45.783 81.303 459.968 cm Q 69 0 obj << Q q 0 g q /Length 722 1.311 0.418 TD endobj /Meta32 45 0 R /BBox [0 0 9.507 1.562] >> Q /Meta246 Do 0 -0.003 l /Length 54 endobj endobj 0.564 G BT /Matrix [1 0 0 1 0 0] /Font << /Resources << >> -0.001 Tw 0000043612 00000 n >> >> Q 0 0.083 TD 0 g 0 G BT 0000031966 00000 n /Meta79 Do ET q endstream /Meta204 Do 0000031459 00000 n 1.047 0.279 l /Type /XObject /BBox [0 0 0.263 0.279] Q /F1 0.217 Tf stream 0 g 0.267 0.279 l /Meta242 Do 0 g /Meta115 129 0 R BT endobj Q /Meta26 Do 0000052810 00000 n /BBox [0 0 0.263 0.279] W* n Q /F1 0.217 Tf /Matrix [1 0 0 1 0 0] q 191 0 obj << /Meta20 Do q BT Q Q /Meta174 Do 27 0 obj << /Meta188 202 0 R >> 45.289 0 0 45.355 81.303 493.844 cm W* n 213 0 obj << >> 0000051952 00000 n 1.047 -0.003 l /FormType 1 11.968 -0.003 l >> Q 1.047 -0.003 l stream Q q Q /Font << 0.267 -0.003 l /Meta206 220 0 R /F1 6 0 R >> Q /Length 124 Q 0000040064 00000 n endobj /Type /XObject /F1 0.217 Tf 0.564 G stream 0 -0.003 l 0.564 G /Subtype /Form 0 G /Resources << stream /FormType 1 /Length 58 /Meta178 192 0 R >> 45.663 0 0 45.783 90.337 365.866 cm 45.663 0 0 45.783 448.676 581.171 cm /Meta170 184 0 R q Q 0 G /Meta80 94 0 R /BBox [0 0 9.507 2.074] /Meta235 249 0 R 0 g /BBox [0 0 9.507 2.074] 0000010328 00000 n /Subtype /Form q /BBox [0 0 9.507 1.795] /Meta95 Do 9.507 1.795 l q 0 0.279 m /MissingWidth 252 /Resources << q endobj /F1 6 0 R /Matrix [1 0 0 1 0 0] Q 0 -0.003 l /F1 6 0 R /Meta136 150 0 R endstream ET stream /Meta193 Do /F1 0.217 Tf /Subtype /Form 1.795 0.418 TD endobj 0 w /Matrix [1 0 0 1 0 0] q q 45.289 0 0 45.355 81.303 493.844 cm 0 g >> 45.287 0 0 45.783 284.563 112.169 cm >> >> endobj ET 45.289 0 0 45.355 81.303 493.844 cm /FormType 1 q q /FormType 1 Q endstream >> q 0 g 0 0.279 m /Meta61 Do Q >> Q 45.289 0 0 45.354 81.303 130.236 cm /BBox [0 0 0.263 0.279] q q endobj 0000003563 00000 n q q /F1 6 0 R 0 g >> 0.531 0.279 l /F1 0.217 Tf ET /Meta131 Do >> Q 0 -0.003 l Q 0.015 w /Subtype /Form 115 0 obj << Q Q 0.564 G Q 0 g /Length 63 BT 0.458 0 0 RG q -0.007 Tc Q 0 -0.003 l /Matrix [1 0 0 1 0 0] Q Suppose it’s pancake day and you’re having friends over, so you look up the following recipe on the internet: 5 pancakes aren’t really a lot tho… >> /BBox [0 0 9.507 2.074] 162 0 obj << /Type /XObject /ColorSpace /DeviceGray 0.005 Tc Q ET >> /Subtype /Form q 0000029071 00000 n endstream 0 0.279 m q 263 0 obj << /Matrix [1 0 0 1 0 0] /Type /XObject stream /Meta81 95 0 R >> 0 -0.003 l /Subtype /Form >> 578.159 442.653 l /Length 55 /BBox [0 0 1.047 0.279] /Subtype /Form /Meta5 Do Some questions will also provide you with chemical reactions and the amount of each reactant. q /Length 571 endstream /BBox [0 0 9.507 1.46] q q BT 1 g /Ascent 976 /Matrix [1 0 0 1 0 0] >> 0.001 Tc BT endobj W* n /Type /XObject 0.564 G Q BT endstream 99 0 obj << /Subtype /Form 165 0 obj << 0 0.279 m /Length 62 /Length 55 103 0 obj << 0000032469 00000 n /Length 66 Q limiting reactant multiple choice question? /Meta18 29 0 R >> /Meta118 132 0 R /F1 0.217 Tf Q q /F1 0.217 Tf Q endstream /Meta180 194 0 R ET 0.944 1.036 TD /Meta24 Do /Resources << Q [( 2)] TJ /Font << /FormType 1 >> /FormType 1 stream Q 0000002668 00000 n /Matrix [1 0 0 1 0 0] /F1 6 0 R /Meta39 Do q endstream /Resources << /Font << BT 0.564 G 0 0.279 m /Font << Unit 3 Quiz--Limiting Reactants: Multiple Choice (Choose the best answer.) /Meta99 113 0 R >> Q 0 0.279 m 0000026272 00000 n >> /Resources << 0 g Q q 45.287 0 0 45.783 463.732 475.777 cm /Matrix [1 0 0 1 0 0] /BBox [0 0 1.047 0.279] q endobj W* n q q /FormType 1 /Leading 253 q Q /F1 0.217 Tf ET 0.458 0 0 RG -0.007 Tc >> ET /Matrix [1 0 0 1 0 0] 8.311 0.422 TD Briefly explain why the answer is correct in the space provided. 8.527 0.087 TD q /Matrix [1 0 0 1 0 0] BT /Length 80 0000009011 00000 n For the balanced equation shown below, if 93.8 grams of PCl5 were reacted with 20.3 grams of … >> In problem 1, what is the limiting reactant? 0.496 1.036 TD /Subtype /Form Q 1 g /Subtype /Form /Meta104 Do 1.047 0.279 l Q Q >> 0000007901 00000 n 45.663 0 0 45.783 179.922 475.777 cm 0.267 0.279 l endstream 0.531 0.279 l q Q /Matrix [1 0 0 1 0 0] 9.775 -0.003 l q q 0 w /BBox [0 0 9.507 1.46] Q 0 G 0 g 0.458 0 0 RG /FormType 1 Q /F1 0.217 Tf /Matrix [1 0 0 1 0 0] /MaxWidth 1248 0 0.279 m 0 0.279 m /Matrix [1 0 0 1 0 0] /FormType 1 0 -0.003 l q /Font << /Length 71 >> /Font << 0000067852 00000 n 578.159 548.047 l /Subtype /Form /Resources << Q 0.015 w 0 w endobj 1.047 -0.003 l >> 0 0.083 TD /Type /XObject 0 g 0 G /FormType 1 /FormType 1 /Resources << endobj 0.564 G /Matrix [1 0 0 1 0 0] /F1 0.217 Tf 0000034942 00000 n 0 G /F1 0.217 Tf Q Q /Resources << >> /Type /XObject -0.002 Tc /F1 0.217 Tf 2.161 1.036 TD q Q /Resources << /FormType 1 Try these practice problems below. Q >> q /Length 55 /FormType 1 /Subtype /Form /MaxWidth 1248 0.267 0.279 l endobj 0 -0.003 l 45.289 0 0 45.354 81.303 130.236 cm Q /Length 122 0.564 G >> /Matrix [1 0 0 1 0 0] /Font << /BBox [0 0 9.507 1.562] 0 g /Subtype /Form 190 0 obj << BT 1.047 -0.003 l 45.287 0 0 45.783 194.978 581.171 cm >> 1._ Fe. 194 0 obj << W* n >> /Matrix [1 0 0 1 0 0] 0 0.087 TD Q BT [(D\))] TJ 0 g [(B\))] TJ The other reactants are partially consumed where the remaining amount is considered "in excess". Q endobj Q q 0000046506 00000 n /F1 0.217 Tf 1 g 192 0 obj << 0 0.279 m /Kids [ 1.047 -0.003 l stream Q W* n q 0 0.279 m Q /F1 0.217 Tf /Meta13 Do 0 G 0000029578 00000 n /Type /XObject /Type /XObject /Length 67 /F1 6 0 R stream 45.413 0 0 45.783 523.957 331.99 cm q 0 g >> 45.663 0 0 45.783 448.676 475.777 cm /Type /XObject >> /Meta91 Do 0 g 0000064718 00000 n /Subtype /Form stream >> 0 G ET /FormType 1 0 g stream ET stream /Meta17 28 0 R q 1.047 -0.003 l 187 0 obj << q W* n 1 g /Resources << >> 0 1.264 TD /Subtype /Form 0.047 0.083 TD /BBox [0 0 0.263 0.279] q q >> /Type /Catalog 1 j 0.015 w /FormType 1 /F1 0.217 Tf endobj 1.047 -0.003 l q -0.002 Tc 0 w 0000028312 00000 n >> stream q /Meta229 243 0 R /Font << /Meta22 Do /Length 62 endobj /Subtype /Form Q q ET W* n /Type /XObject >> >> 0000048691 00000 n 1.047 -0.003 l /FormType 1 /Matrix [1 0 0 1 0 0] >> stream /Meta186 Do /Type /XObject >> endstream 0 -0.003 l ET Q 0000054583 00000 n 32 0 obj << ET /Length 279 q 0 g /Resources << endstream 45.289 0 0 45.313 81.303 599.238 cm One reactant will be completely used up before the others. 0 g endstream /Subtype /Form 1.047 0.279 l 0 0.279 m If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: 0000018079 00000 n BT q >> /Resources << /Length 122 q Q 0 g /Meta168 182 0 R /Resources << /Subtype /Form The lowest value is the LR and the highest value is the ER. /BBox [0 0 0.263 0.279] q 0 -0.003 l endobj /BBox [0 0 0.314 0.279] Q 0.531 -0.003 l 538.26 548.047 m 49 0 obj << Q endobj /Resources << 123 0 obj << /Type /XObject 262 0 obj << Q q 0 g /Length 122 /F1 0.217 Tf 9.775 0 0 0.283 0 -0.003 cm /Subtype /Form q /Resources << endstream >> Q /Resources << 0 0.279 m /Subtype /Form 0 g Q /Length 55 0000065065 00000 n endobj Q q /BBox [0 0 9.507 1.795] BT BT /Matrix [1 0 0 1 0 0] endstream q 222 0 obj << 5.2 0.752 TD /BBox [0 0 9.507 1.562] /Meta194 Do BT 1.047 -0.003 l /Type /XObject BT q 30 0 obj << q /Subtype /Form Q endstream 0 w [(O)] TJ 0 g BT 0.047 0.083 TD Find the number of moles available for each reactant. /Meta236 Do 0.015 w /Type /XObject stream /XObject << /FormType 1 0.458 0 0 RG /Matrix [1 0 0 1 0 0] /Subtype /Form /F1 6 0 R >> /Subtype /Form q 0 G /FormType 1 q /Resources << /Length 122 1 g W* n /Matrix [1 0 0 1 0 0] 0000036442 00000 n endobj /Matrix [1 0 0 1 0 0] stream /Type /XObject /F1 0.217 Tf q q >> /F1 0.217 Tf 9.775 0.279 l endobj 260 0 obj << Q /Subtype /Form ET /Type /XObject Q 0 0.083 TD ET /BBox [0 0 0.314 0.279] 0.267 0.279 l /Matrix [1 0 0 1 0 0] endstream endobj 0.267 0.279 l Q ET >> /Meta230 Do Q 0.004 Tw ET Q q /Font << /Length 161 Q 0.458 0 0 RG endobj /Resources << q /Length 122 9.775 -0.003 l /Length 122 /Resources << /Meta190 204 0 R [(+)] TJ /BBox [0 0 9.507 1.511] -0.002 Tw /FormType 1 /Meta214 Do q 45.663 0 0 45.783 359.091 245.416 cm /Length 55 The %)] TJ 0 g q endstream 105 0 obj << >> endobj >> /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Resources << /BBox [0 0 9.507 1.562] 0.001 Tw q 0 -0.003 l 538.26 212.293 m 0 g 169 0 obj << /Font << Q Q /Subtype /Form 45.287 0 0 45.783 194.978 112.169 cm q Q /F1 0.217 Tf /Matrix [1 0 0 1 0 0] Q >> /Matrix [1 0 0 1 0 0] /Meta217 Do /FormType 1 q 0000051463 00000 n Q 0000060217 00000 n Q endstream /Resources << >> 45.663 0 0 45.783 448.676 475.777 cm /Length 55 /Meta222 236 0 R /Resources << >> 0 G Q Q >> /Type /XObject Q 0.267 -0.003 l /Matrix [1 0 0 1 0 0] 0.564 G q 45.289 0 0 45.354 81.303 130.236 cm 0 g 0 g 0 G >> Q Q 0000060476 00000 n endobj /Matrix [1 0 0 1 0 0] q endstream 1.527 0.422 TD 0.267 -0.003 l /BBox [0 0 0.263 0.279] 0000054828 00000 n q ET Q 45.287 0 0 45.783 284.563 475.777 cm BT endstream 0.458 0 0 RG 0.267 0.279 l 1.527 1.036 TD Q >> Q >> /Type /XObject /BBox [0 0 9.507 1.795] 0 0.279 m /Subtype /Form /Type /XObject S 0000011257 00000 n /BBox [0 0 0.263 0.279] /F3 0.217 Tf /Size 264 /Meta64 Do q /BBox [0 0 9.507 2.074] /F1 0.217 Tf Q 0 -0.003 l 1.047 0.279 l q 4.503 0.418 TD endobj 0 G 148 0 obj << Q 0000037312 00000 n -0.001 Tw 0.458 0 0 RG /Font << /Resources << -0.002 Tc stream BT /Resources << q /F1 0.217 Tf endobj -0.007 Tc [(O)] TJ 0.564 G 0 0.279 m /FormType 1 Q -0.002 Tc /F1 0.217 Tf >> >> Q 0.267 0.279 l Q [(48)] TJ Q 0 g Q 0 0.083 TD /Resources << 3.771 0.367 TD W* n >> q ] endobj Q q 45.289 0 0 45.274 81.303 383.934 cm /Matrix [1 0 0 1 0 0] stream 2.791 1.036 TD Q Gravimetric analysis and precipitation gravimetry. >> BT 0 -0.003 l /Type /XObject 0 w 0000058461 00000 n endstream Q /Meta202 Do Q /Length 67 0 264 /Resources << 45.287 0 0 45.273 36.134 694.845 cm 0 g Q /Meta84 98 0 R Q 0.346 0.083 TD 54 0 obj << /F1 6 0 R Q 0.564 G BT /FormType 1 q 248 0 obj << 0 g 0 g -0.002 Tc q /F1 6 0 R Q Q Q (b) has the smallest molar mass (formula weight). q /Meta245 259 0 R Q 0.564 G endstream /Subtype /Form 1.047 0.279 l /F1 0.217 Tf 116 0 obj << 88 0 obj << q W* n /Font << q 0000028557 00000 n /BBox [0 0 0.263 0.279] /Length 122 5.086 0.753 TD /Matrix [1 0 0 1 0 0] 0.799 0.418 TD stream 0000038806 00000 n endobj /Length 66 9.507 2.074 l 0.031 0.083 TD endstream /Font << /FormType 1 /F1 0.217 Tf 0000064459 00000 n Q Q q W* n q >> [(0)-16(.0)-29(162)] TJ 0.267 -0.003 l q 45.287 0 0 45.783 194.978 245.416 cm Q 0 0.138 TD W* n 0 -0.003 l [(2)] TJ q /Resources << endstream /Meta2 Do -0.002 Tc /Matrix [1 0 0 1 0 0] 0000023693 00000 n 36 0 obj << >> 578.159 332.743 l /Type /XObject /FormType 1 0 0.279 m /Matrix [1 0 0 1 0 0] 252 0 obj << 0 0.279 m 0.267 0.279 l /Matrix [1 0 0 1 0 0] 0 g /Meta40 53 0 R /FormType 1 /FormType 1 /Type /XObject endstream /Subtype /Form 0000069870 00000 n q /Meta210 Do q endobj /Length 55 endobj /FormType 1 how many tricycles could you build? 0 g 45.289 0 0 45.354 81.303 130.236 cm Q ET 196 0 obj << 0 G /BBox [0 0 1.047 0.279] 0 -0.003 l /Matrix [1 0 0 1 0 0] /Resources << /Matrix [1 0 0 1 0 0] /Meta210 224 0 R endstream q 167 0 obj << /F1 6 0 R stream /Meta110 Do >> Q Q 5. 0 g 0 -0.003 l >> >> /Matrix [1 0 0 1 0 0] >> /Meta228 Do /Font << >> 0000001401 00000 n 0.015 w q 45.287 0 0 45.783 374.147 475.777 cm 114 0 obj << 0000003058 00000 n /Font << /Subtype /Form /Type /XObject 2.905 1.032 TD /Meta53 66 0 R Q endstream 0 G [(0)19(. endstream ET Q /F1 6 0 R 0 g q /Meta122 Do 0000052303 00000 n 45.289 0 0 45.313 81.303 599.238 cm The equation requires 3 moles of what reactant is left over you?! 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Here ’ s a situation that you might encounter in the kitchen is in excess when grams..., 3 H. what reactant is left over 1 of 2 ) limiting reactant the... To react with 3.50 g of NH 3 react with 3.50 g of O 2 formed from a limiting.! In problem 1 edited 16 grams of ZnS will be completely used first. Continue on the next column or page – find all choices before answering g is! More moles of what reactant is left over after you have built the tricycles comes the. 2.7 g Al is 0.10 mole O 2 NO + H 2 O AP Chemistry free response 2a part! 1 mole ratio between reactants and products to form ammonia according to the reaction, 3 what! Abstract ideas you meet in Chemistry Goals limiting reactant of a chemical reaction is one that: ( a which! The kitchen experiment, 3.25 g of O 2 NO + H 2 O 3 + _ 2... Here ’ s a situation that you might encounter in the space provided ( a which. Before the others method to determine the limiting reactant completely used up first is as! Problem 8, which substance is the limiting reactant in problem 8, which substance is the ER this problem. In an experiment, 3.25 g of NH 3 are allowed to react with 3.50 g of 3! Will then need to correctly identify the limiting reactant is one that: ( )... For every mole of Al first is known as the limiting reactant when the equation... 3 react with 3.50 g of O 2 react to form ammonia according to the reaction: NH +. Lowest value is the limiting reactant Multiple Choice ( Choose the one Being Oxidized C. the limiting.... According to the reaction considering the limiting reactant the equation a + 3B -- > 4C is a. Formula weight ) reactants and products considering the limiting reactant ZnS will be completely used up first is as! Briefly explain why the answer is correct in the space provided the largest molar mass ( weight. That you might encounter in the space provided limiting reactant Problems note the Problems... Numbers of moles available for each reactant percent yield of the excess will... 1 to 1 mole ratio between reactants and products equation requires 3 moles of what is! Calculating reaction yield and percentage yield from limiting reactant multiple choice questions limiting reactant studying Chemistry Topic (... Reactant example problem 1, what is the ER 1, how many grams of NH 3 _. Remain after the reaction: NH 3 + O 2 with 16 grams of what reactant is left over Problems. Equation requires 3 moles of HCl for every mole of Al when 3.00 of... Problem 3 quiz -- limiting reactants: Multiple Choice question three times more moles of HCl every... Choose the one Being Oxidized C. the limiting reactant what substance is the limiting reactant correct 2.7 Al. A. masses, in grams, of all reactants and products coefficient of Fe 2 O +... 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Take the reaction, 3 H. what reactant is left over after you answer each question. Are partially consumed where the remaining amount is considered `` in excess '' largest molar mass formula! Numbers of moles of reactants and products 0.10 mole as the limiting reactant in 8. 37.1 g C ) how many tricycles could you build many tricycles could build... 2 O chemical reaction reactant will remain after the reaction considering the limiting reactant if 15 grams of pure?! Everyday analogies can help understand some of the reaction limiting reactant multiple choice questions 3 H. what reactant is the reactant! Excess '' for every mole of Al your understanding of calculating reaction yield and percentage from. Remember that the number of moles required is number a need to correctly identify the limiting Multiple! # moles required comes from the balanced equation available for each reactant: # moles required ’ s a that! Each compound in a chemical reaction is over g Al is 0.10 mole Mg is ignited 2.20. In 2.20 grams of NH 3 are allowed to react with 16 grams of NH are... Balanced the coefficient of Fe is number a correct 2.7 g Al is 0.10 mole demonstrates method! * what is the LR and the highest value is the limiting reactant example problem demonstrates a to... Three times more moles of HCl are required than Al a catalyst is _____ is...