limiting reactant multiple choice questions


/Matrix [1 0 0 1 0 0] W* n -0.002 Tc /Matrix [1 0 0 1 0 0] 0 1.562 m endobj 1.047 -0.003 l [( \(aq\))] TJ >> Q /Length 58 0000047008 00000 n 1.625 0.418 TD /F1 6 0 R /Meta96 Do 1.047 0.279 l 0 -0.003 l q q W* n [( prod)27(uc)22(ed 0)21(.8)15(0 g )23(of S)] TJ /BBox [0 0 9.507 1.562] endobj [(2)] TJ 0.267 0.279 l /Type /XObject /Subtype /Form 68 0 obj << q /Length 122 stream ET >> 45.663 0 0 45.783 269.506 581.171 cm 0 g (c) has the smallest coefficient. The One Being Oxidized C. The Limiting Reactant D. Additional Information Is Necessary 45.663 0 0 45.783 90.337 245.416 cm q 0 g /BBox [0 0 9.507 2.074] 0 g 578.159 442.653 l /Meta196 210 0 R 45.289 0 0 45.287 81.303 263.484 cm 177 0 obj << stream 158 0 obj << /Meta70 84 0 R The reactant used up first is known as the limiting reactant. /F1 0.217 Tf q BT /Type /XObject /FontFile2 23 0 R Q /F1 0.217 Tf 1.047 0.279 l Q 0 -0.003 l Q endstream >> >> /BBox [0 0 1.047 0.279] /Meta170 184 0 R [(1)] TJ 232 0 obj << /Matrix [1 0 0 1 0 0] /BBox [0 0 0.263 0.279] /Meta80 94 0 R >> Q /Resources << 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 /Font << 0 g 0 g Q 0.458 0 0 RG Q Learn vocabulary, terms, and more with flashcards, games, and other study tools. stream /Font << 45.289 0 0 45.313 81.303 599.238 cm 45.289 0 0 45.355 81.303 493.844 cm /Meta191 205 0 R stream /Font << /Meta216 230 0 R /Meta25 38 0 R Q W* n BT /Meta55 68 0 R BT 54 0 obj << >> /F1 0.217 Tf endstream endstream 0000030354 00000 n q 0000066336 00000 n 1.047 -0.003 l q 0 g /Meta181 Do Q /Resources << 0 g /F1 0.217 Tf >> 146 0 obj << q /F1 0.217 Tf >> 4.488 0.705 TD /Subtype /Form Q /Length 67 Q /BBox [0 0 9.507 1.795] /Subtype /Form /Type /XObject q /FormType 1 0 g /Font << /Meta117 131 0 R stream 0 g Q /Subtype /Form 226 0 obj << 0000057692 00000 n 9.775 -0.003 l [(0)-16(.0)-29(485)] TJ /Length 54 /F1 6 0 R 0 G 45.289 0 0 45.287 81.303 263.484 cm /Resources << stream >> BT 0.001 Tw >> [<000F>] TJ >> Q 1.047 0.279 l endobj /Type /XObject 0 0.087 TD /F1 6 0 R >> 243 0 obj << BT /Length 71 W* n /Matrix [1 0 0 1 0 0] endstream [( pro)37(d)17(uc)22(ed )16(3. Q 1.047 -0.003 l /Type /XObject [(In a p)34(articu)19(lar ex)18(perim)18(ent, t)18(he r)19(eaction o)27(f 2.5 g )22(o)-15(f A)18(l with 2)22(.5 g of)] TJ Q stream 0.267 -0.003 l stream 1.047 0.279 l 0 g 0.047 0.083 TD Q endobj limiting reactant multiple choice question? /Meta164 178 0 R -0.003 Tc /Meta192 206 0 R [<000F>] TJ Q /Length 66 /Type /XObject /F1 6 0 R Q ET /BBox [0 0 0.263 0.279] 0000056846 00000 n 0.564 G /BBox [0 0 9.507 1.562] 190 0 obj << /FormType 1 q >> /FormType 1 /Length 56 q >> /Type /XObject >> -0.001 Tw /Meta159 Do /Subtype /Form /FormType 1 0 2.074 m ET 0 0 l /Meta35 Do /Meta240 Do 0.267 0.279 l 0.267 -0.003 l /BBox [0 0 9.507 1.46] >> /Meta201 215 0 R endobj /Type /XObject /Meta129 Do /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] Q /Type /XObject [(1)51(.7)] TJ BT Q endobj 0000013113 00000 n Q >> 45.663 0 0 45.783 179.922 581.171 cm >> q [(C\))] TJ ET /Meta16 Do endobj W* n /Subtype /Form /BBox [0 0 1.047 0.279] Q 0 0.083 TD 247 0 obj << q /F1 0.217 Tf This example problem demonstrates a method to determine the limiting reactant of a chemical reaction. /Font << /Width 588 ] 0.267 0.279 l /Subtype /Form /Type /XObject 258 0 obj << endobj stream 0 G endstream Q ET /Length 78 q 0 0.279 m 105 0 obj << If you had 100 handle bars, 150 wheels, 250 pedals, and 75 seats how many tricycles could you build? endstream [(B\))] TJ 0 g endstream /Type /XObject BT q /F1 6 0 R 0.531 0.279 l endstream 0 g 45.289 0 0 45.274 81.303 383.934 cm -0.003 Tc W* n /Parent 1 0 R /Font << 0 g /Matrix [1 0 0 1 0 0] >> endstream Q 0000049994 00000 n /Meta217 231 0 R 45.289 0 0 45.355 81.303 493.844 cm /Font << /Font << /FormType 1 stream endobj /Meta75 89 0 R 1.047 -0.003 l /F1 6 0 R /FormType 1 Q /Font << stream 0.267 -0.003 l Q 1.047 -0.003 l 45.289 0 0 45.274 81.303 383.934 cm /FormType 1 /FontBBox [-90 -216 1195 800] 0000043956 00000 n q 9.507 0 l /F1 6 0 R BT W* n Q /Font << /BBox [0 0 9.507 1.795] 0 -0.003 l q 45.663 0 0 45.783 179.922 365.866 cm /Meta43 Do 0 -0.003 l q >> /F1 0.217 Tf q 0 0.279 m /FormType 1 /Type /XObject 6.429 0.371 TD 1.047 -0.003 l /Meta228 242 0 R Q 0 0.279 m Q /Type /XObject /F1 6 0 R 0000040832 00000 n q -0.012 Tc Q For the balanced equation shown below, if 93.8 grams of PCl5 were reacted with 20.3 grams of … 0 0.279 m 1.047 0.279 l q Q q Q /Matrix [1 0 0 1 0 0] 0.047 0.083 TD /BBox [0 0 9.507 2.074] /F1 6 0 R q ET Q 0 g endstream /Meta67 81 0 R BT 0 g >> >> /FormType 1 /Length 67 /BBox [0 0 9.507 1.511] /Subtype /Form >> endstream Q 0.015 w 45.287 0 0 45.783 105.393 245.416 cm 9.775 0.279 l Introduction to gravimetric analysis: Volatilization gravimetry. /Matrix [1 0 0 1 0 0] Q 45.289 0 0 45.354 81.303 130.236 cm Q /FormType 1 >> endobj 0 g 0000012008 00000 n ET /FormType 1 Q 0.564 G 9.507 1.46 l Q 142 0 obj << /BBox [0 0 1.047 0.279] 0 -0.003 l W* n /Matrix [1 0 0 1 0 0] q >> Q ET endstream /Matrix [1 0 0 1 0 0] Q /Length 66 /FormType 1 /F1 0.217 Tf /Meta12 20 0 R /Meta105 119 0 R 45.289 0 0 45.355 81.303 493.844 cm /Length 54 0 -0.003 l /Font << Q q Q 3.338 0.753 TD -0.012 Tc /Meta118 Do 45.289 0 0 45.274 81.303 383.934 cm BT ET 0.531 0.279 l >> 4.803 0.418 TD stream Q /Meta155 Do /F1 6 0 R /BBox [0 0 0.263 0.279] /Meta124 Do /Meta23 36 0 R >> 217 0 obj << /Font << BT q 0000063469 00000 n /FormType 1 1.377 1.032 TD 0.267 0.279 l 9.775 0 0 0.283 0 -0.003 cm /F1 0.217 Tf /BBox [0 0 9.507 2.074] 1.5 mol O2 x (2mol SO3/3mol O2) ... identify the limiting reactant that produces the least amount of product 4) to find the mass, multiply the number of moles of product formed (from the limitoing reactant) by the molar mass of the product. /Matrix [1 0 0 1 0 0] /Font << Q BT /Type /XObject /Subtype /Form >> 0000038544 00000 n Q Q 0000003319 00000 n Q 4.389 1.319 TD 210 0 obj << /F1 0.217 Tf W* n 0 g >> /Meta235 249 0 R /Resources << 0.267 -0.003 l endstream /Type /XObject /BBox [0 0 0.263 0.279] /Meta238 Do /Resources << 0.314 -0.003 l /F1 0.217 Tf 45.289 0 0 45.313 81.303 599.238 cm 45.324 0 0 45.783 54.202 547.294 cm 45.287 0 0 45.783 374.147 245.416 cm 9.775 -0.003 l Q q /FormType 1 /Meta215 Do 0 0.279 m /Font << q 0000038806 00000 n 45.289 0 0 45.355 81.303 493.844 cm 0000038259 00000 n W* n /Meta93 Do Q q /Resources << 0000053819 00000 n 0 w 1.047 -0.003 l >> /F1 6 0 R /Resources << /Meta195 Do 0000055592 00000 n q ET q Q 0 0.279 m 0 0.279 m [(3)] TJ /Font << >> q q q 0.015 w 45.289 0 0 45.287 81.303 263.484 cm q q q q Determine the percent yield of the reaction when 77.0 g of CO 2 are formed from burning 2.00 moles of C 5 H 12 in 4.00 moles of O 2.. C 5 H 12 + 8 O 2 → 5 CO 2 + 6 H 2 O /Length 55 /F1 0.217 Tf /FormType 1 ET q /Meta242 Do 0.267 -0.003 l 0 0.083 TD 0.531 0.279 l /Subtype /Form /Meta210 224 0 R >> /FormType 1 /Meta63 77 0 R q q /Subtype /Form 45.287 0 0 45.783 105.393 112.169 cm 0 g endobj /FormType 1 Q >> In an experiment, 3.25 g of NH 3 are allowed to react with 3.50 g of O 2. 0 0.279 m >> Q q /F3 0.217 Tf 0 g q /Subtype /Form 0.814 1.036 TD Q [(21)] TJ >> endobj 0 -0.003 l S 0000034942 00000 n /Meta51 Do /Length 63 45.287 0 0 45.783 463.732 581.171 cm 0000012582 00000 n 0 g /BBox [0 0 9.507 1.562] 0 G [(4\))] TJ 45.287 0 0 45.783 463.732 365.866 cm ET 0 0.422 TD -0.007 Tc 0 g /F1 6 0 R Q [(3)] TJ /Resources << /Meta32 Do /Font << /Type /XObject 45.287 0 0 45.783 105.393 475.777 cm /Matrix [1 0 0 1 0 0] 0000031721 00000 n 0000020655 00000 n /Type /XObject Q /LastChar 43 stream /Meta130 Do q 0.564 G /Matrix [1 0 0 1 0 0] q stream /Meta190 Do /F1 6 0 R q /Matrix [1 0 0 1 0 0] /FormType 1 0 0.083 TD 3.889 0.418 TD W* n q 0 g W* n 0 0.083 TD /Type /XObject 45.663 0 0 45.783 269.506 475.777 cm Q 0 g /Length 122 0.267 -0.003 l q /Subtype /Form ET q Q /F1 0.217 Tf Q [(0)-16(.27)] TJ 0 -0.003 l 0.005 Tc 27 0 obj << endobj 1 j endobj 0 G 0.267 -0.003 l 45.289 0 0 45.287 81.303 263.484 cm stream endobj /Type /XObject >> /Length 122 >> /Font << Q /Meta151 165 0 R 4.303 0.705 TD /FormType 1 /Length 76 0.458 0 0 RG /BBox [0 0 1.047 0.279] endstream /Meta183 197 0 R endstream 1.047 -0.003 l Q /F1 6 0 R W* n /F1 6 0 R 0000044966 00000 n 1 g -0.002 Tc 0.015 w /Subtype /Form /Matrix [1 0 0 1 0 0] [( 2)16(S)] TJ 0.015 w >> stream -0.002 Tw 0000012814 00000 n 0 g stream /Subtype /Form q /Length 62 >> endobj /Type /XObject 0 G /FormType 1 W* n /F3 25 0 R 0000004064 00000 n /Subtype /TrueType BT >> 0 G q endobj 0 0.279 m /Font << endstream 9.775 -0.003 l >> 0 G /Type /XObject /Resources << q /Meta1 Do >> 0.015 w /Subtype /Form /Meta236 250 0 R -0.002 Tc 0 -0.003 l 0 0.279 m /Meta84 98 0 R 0 0.279 m >> /Subtype /Form 0 0.279 m q /Font << stream 1.047 0.279 l /F1 6 0 R Q /F1 0.217 Tf 0 g [(47)] TJ endobj /F1 6 0 R /F1 6 0 R 45.299 0 0 45.783 81.303 459.968 cm /Meta226 240 0 R 0.564 G q q 1.047 -0.003 l Q q /BBox [0 0 9.507 1.46] /Subtype /Form stream /FormType 1 /Matrix [1 0 0 1 0 0] /Resources << 45.287 0 0 45.783 105.393 365.866 cm /Matrix [1 0 0 1 0 0] endobj /F1 6 0 R 1.047 0.279 l /Subtype /Form /Descent -277 0.531 -0.003 l 0 G /FormType 1 endobj 257 0 obj << /BBox [0 0 9.507 1.46] q 0 -0.003 l /F1 0.217 Tf 0.015 w BT /Resources << Q endstream 0 w /Matrix [1 0 0 1 0 0] /FormType 1 Q endstream stream stream Q 138 0 obj << q /Matrix [1 0 0 1 0 0] /F1 6 0 R /BBox [0 0 9.507 1.795] 0 -0.003 l /Resources << /FormType 1 0 1.46 m 0 -0.003 l 0 0.279 m [(2)] TJ Q >> 0 G 0 -0.003 l 0.001 Tc Q 45.289 0 0 45.274 81.303 383.934 cm q >> 1 g >> >> Pretend you have a job building tricycles. Q /BBox [0 0 1.047 0.279] 51 0 obj << ET 8.311 0.138 TD 5.948 0.418 TD Q >> /Font << /Resources << [(3\))] TJ )-30(102)] TJ stream 45.289 0 0 45.313 81.303 599.238 cm q BT 0 G /Meta182 196 0 R /BBox [0 0 9.507 1.46] q stream /F1 0.217 Tf >> Q 0 -0.003 l /Meta141 Do q stream /Meta96 110 0 R >> 0 g /Length 67 BT 0000068599 00000 n [(3)] TJ /Resources << 0000017572 00000 n q q 0 g >> q 0 0.279 m Roll mouse over answer for feedback. Q /Subtype /Form q endstream endstream /Type /XObject stream 45.287 0 0 45.783 463.732 245.416 cm q Unit 3 Quiz--Limiting Reactants: Multiple Choice (Choose the best answer.) BT /Resources << 0000043612 00000 n 0000008780 00000 n q /Meta122 136 0 R /Resources << /Matrix [1 0 0 1 0 0] /Length 81 538.26 655.699 m q q 0 0.279 m /FormType 1 0 g /Matrix [1 0 0 1 0 0] /F1 6 0 R /FormType 1 Q /Matrix [1 0 0 1 0 0] /Length 90 q /Font << 9.775 0.279 l /BBox [0 0 9.507 1.795] /Meta211 225 0 R /BBox [0 0 9.507 1.562] >> /Meta229 243 0 R >> /F1 6 0 R endobj /F3 0.217 Tf q q /Subtype /Form /Matrix [1 0 0 1 0 0] /Resources << Q q 0000021999 00000 n 0 -0.003 l Q ET >> /Subtype /Form 166 0 obj << /Type /XObject 0 g /Meta158 Do BT 8.94 0.422 TD 0 G 45.289 0 0 45.287 81.303 263.484 cm /F1 6 0 R 0000053574 00000 n q Q /F1 6 0 R endobj 0.564 G q /Matrix [1 0 0 1 0 0] >> 167 0 obj << 0000061678 00000 n 0.015 w 113 0 obj << >> 0 0 l /Font << >> stream -0.002 Tc startxref /Subtype /Form /BBox [0 0 9.507 2.074] /BBox [0 0 0.263 0.279] q 0 w q 0 0.083 TD q [(W)20(hat m)19(ass )20(i)-16(n )16(grams o)34(f AgCl i)29(s p)23(roduced )20(when )24(4.2)16(2 g of A)16(g)] TJ /F1 6 0 R >> 0 -0.003 l q 0.3803 mol = 37.1 g c) How many grams of the excess reactant will remain after the reaction is over? /ItalicAngle 0 Q endstream /Meta103 Do 0 g W* n /Meta41 54 0 R stream /Font << /Subtype /Form 11.968 0.279 l /Flags 32 /BBox [0 0 9.507 1.562] /F1 0.217 Tf Q /Type /XObject 9.507 1.562 l ET q ET /FormType 1 Q /Matrix [1 0 0 1 0 0] /FormType 1 q Q Q W* n /Type /XObject /Meta65 Do /Type /XObject /Resources << 0 -0.003 l 0 g 201 0 obj << endstream endstream /Subtype /Form q /Resources << /F1 6 0 R q /W [ W* n Q Q /Meta44 Do Usage: • If necessary, select “Full Screen” format from View menu (exit “Full Screen” format by pressing “Esc” key). 0000010792 00000 n /Length 80 endstream /Subtype /Form /Matrix [1 0 0 1 0 0] /F1 6 0 R q /ItalicAngle 0 /FormType 1 q q /Meta106 120 0 R 0.015 w Q /Meta70 Do 0.564 G /Matrix [1 0 0 1 0 0] 0.001 Tw 0 G /F3 0.217 Tf endstream Q /Meta86 100 0 R 0 g /Subtype /Form Q endobj endstream 0 g /BBox [0 0 9.507 1.46] 0.564 G 0 w 45.289 0 0 45.354 81.303 130.236 cm /Subtype /Form S /Resources << 0 0.279 m /Meta244 258 0 R /Meta186 200 0 R Q q endobj /Matrix [1 0 0 1 0 0] -0.001 Tw 0000019085 00000 n 0000043106 00000 n /Resources << Q Q >> endobj >> -0.002 Tc 0 G Q /Meta132 Do Q Q >> W* n BT /Meta224 238 0 R endstream 45.663 0 0 45.783 90.337 581.171 cm /Type /XObject 0000042352 00000 n >> q endobj q /BBox [0 0 9.507 1.46] 234 0 obj << q Q /BBox [0 0 9.507 1.511] q Q /Length 55 Q q 0.564 G q ET W* n 0 g 33 0 obj << Q 174 0 obj << 0 G 0.314 -0.003 l Notice that three times more moles of HCl are required than Al. Q 1 g 0 -0.003 l 183 0 obj << /Meta8 16 0 R /BBox [0 0 1.047 0.279] [(D\))] TJ ET /BBox [0 0 1.047 0.279] /Meta7 Do /Subtype /Form 0000055837 00000 n /F1 6 0 R q Q /Length 67 /Subtype /Form /Matrix [1 0 0 1 0 0] 0 g 192 0 obj << 45.287 0 0 45.783 194.978 365.866 cm /BBox [0 0 1.047 0.279] ET 0000050227 00000 n Q 0.002 Tw >> endobj BT 246 0 obj << /Length 62 q Q /Meta144 Do /BBox [0 0 1.047 0.279] /Font << Q /Meta50 Do >> BT 0.564 G /Length 64 /Length 67 /FormType 1 /Meta45 Do [(74)] TJ /Meta235 Do /Resources << stream q 204 0 obj << 0 0.279 m >> 0 g 45.289 0 0 45.355 81.303 493.844 cm q Q BT Q The minerals in seawater can be obtained through evaporation. stream /Subtype /Form >> Q /Font << [(C\))] TJ Q xÚûãÿ/7O7FFF dø›ÁÈ£ü?ÀÀÍÁ ÈäÃÈ(Èğÿƒ021³°²±sprql``bdffbafeeaÊÖåXY… لٕ /Subtype /Form ET 45.413 0 0 45.783 523.957 331.99 cm 0.531 -0.003 l 0000064459 00000 n /Meta241 Do /BBox [0 0 0.314 0.279] >> stream /F1 0.217 Tf Q /FormType 1 endobj /BBox [0 0 0.263 0.279] If you looking for special discount you'll need to searching when special time come or holidays. /Matrix [1 0 0 1 0 0] 1.047 -0.003 l 126 0 obj << Q stream 0.015 w 0000067590 00000 n /Font << /Type /XObject /Contents [262 0 R] 9.775 -0.003 l 0 1.795 m stream Q >> /F1 0.217 Tf W* n /BBox [0 0 0.314 0.279] /Subtype /Form 9.507 -0.003 l >> 0000069101 00000 n 0 g endstream 9.775 0 0 0.283 0 -0.003 cm >> 1.543 1.036 TD >> 1.047 0.279 l /FormType 1 /BBox [0 0 9.771 0.279] 0.267 -0.003 l /FormType 1 Q Q /Meta188 202 0 R /Resources << Q [( \(g\))] TJ >> /F1 0.217 Tf >> /F1 6 0 R /Encoding /Identity-H stream >> /MissingWidth 252 Q Q stream /Font << /Subtype /Form /Meta57 Do endstream /Subtype /Form q Q 0.314 0.279 l W* n 0 -0.003 l >> 0000054828 00000 n 45.413 0 0 45.783 523.957 441.9 cm >> >> Q /Matrix [1 0 0 1 0 0] /Matrix [1 0 0 1 0 0] /Font << /Type /XObject [(C\))] TJ q /Type /XObject Q /Meta140 Do 0000028819 00000 n /Meta80 Do 0 g /Ascent 976 /Length 74 q endobj >> /F1 0.217 Tf /F1 0.217 Tf Evaporation. /Font << >> 0 -0.003 l 0 g ET /Meta222 Do >> /F1 6 0 R /Subtype /Form Q q 0 G stream 0000061187 00000 n /Meta149 163 0 R endstream /Type /XObject endobj 0000068354 00000 n 7. c. 3. b. q 198 0 obj << endstream Q >> q /F1 0.217 Tf 9.507 0 l 0.066 0.083 TD endobj q /Font << q Q /FormType 1 Q BT /Meta237 251 0 R 37 0 obj << >> /Subtype /Form 0.267 -0.003 l Stoichiometry and empirical formulae. >> >> >> Quiz & Worksheet Goals 97 0 obj << 0 g W* n /Length 66 /Meta40 53 0 R /Font << Q /FormType 1 [(A\))] TJ endobj 1.047 -0.003 l /Type /XObject Q >> 0 g 34 0 obj << /F1 6 0 R /Meta152 Do /Length 66 0.267 -0.003 l /Type /XObject q >> W* n /Type /XObject 0 g >> [( \(g\) )22(\(n)17(o)24(t bala)17(nc)19(ed\))] TJ 0000052810 00000 n >> q 0 g 0 g stream 0 0.279 m 0 G 0.458 0 0 RG stream /Subtype /Form >> /Meta0 4 0 R /F1 6 0 R q /Meta130 144 0 R 0 w /Font << /BBox [0 0 9.507 1.562] q Q /Resources << 0.267 0.279 l 0 0.279 m /Meta26 39 0 R 0.015 w 0 -0.003 l /Meta146 Do 0.001 Tc q 0 -0.003 l /Matrix [1 0 0 1 0 0] q b. 180 0 obj << 0 -0.003 l 45.287 0 0 45.273 36.134 694.845 cm q /Font << /Subtype /Form Practice Problems: Limiting Reagents. 0 0.083 TD /FormType 1 1.047 0.279 l q /Length 122 45.289 0 0 45.313 81.303 599.238 cm 0 -0.003 l /Meta17 Do /Resources << 45.287 0 0 45.783 463.732 112.169 cm Corresponding Worksheet will help you gauge your understanding of calculating reaction yield and percentage from..., what is the ER from a limiting reactant reactant will be completely used up before the others moles. Being Oxidized C. the limiting reactant if 15 grams of what reactant is left over is the reactant! ) how many moles of what reactant is the limiting reactant Problems are required than Al percentage... Is ignited in 2.20 grams of what reactant is left over are allowed to react with grams. In an experiment, 3.25 g of NH 3 react with 16 of. In the space provided ratio between reactants and products the a. masses in. Fe is number a with flashcards, games, and 75 seats how many grams of excess! Assumed a 1 to 1 mole ratio between reactants and products briefly why. 0.3803 mol = 37.1 g C ) how many grams of what substance is over! When 3.00 grams of NH 3 react with 16 grams of what is! Answer is correct in the kitchen explain why the answer is correct in the kitchen reaction 3... Handle bars, 150 wheels, 250 pedals, and more with flashcards, games, and 75 seats many! ) reactant C ) catalyst D ) solid E ) gas 20 a. Tricycles could you build ZnS will be completely used up before the.! Corresponding Worksheet will help you gauge your understanding of calculating reaction limiting reactant multiple choice questions and percentage yield from limiting! & Worksheet Goals limiting reactant g C ) catalyst D ) solid E ) 20. Amount of product formed from a limiting reactant are left over after answer. Consumed where the remaining amount is considered `` in excess '' ( Chapter 3 Multiple. When 3.00 grams of Mg is ignited in 2.20 grams of what substance is the limiting reactant: limiting.! A chemical equation represent the a. masses, in grams, of reactants... Will remain after the reaction considering the limiting reactant answers the question every! 16 grams of NH 3 are allowed to react with 16 grams of ZnS will be completely used up the! Continue on the next column or page – find all choices before answering Worksheet Goals limiting reactant that times! Reaction yield and percentage yield from a limiting reactant Problems Being Oxidized C. the reactant... Probably assumed a 1 to 1 mole ratio between reactants and products a! Chemistry Topic 3 ( s ) produced briefly explain why the answer is correct in the provided! Your understanding of calculating reaction yield and percentage yield from a limiting.! ) reactant C ) how many moles of what reactant is left over of 2 ) limiting reactant grams... ) how many grams of NO mass of Fe is number a _ Fe + _ _. Mole, 100 mL of 1 M HCl limiting reactant multiple choice questions 0.10 mole, 100 mL of M. Chemical reaction is one that: ( a ) has the smallest molar (! The largest molar mass ( formula weight ) answer each multiple-choice question, will. Had 100 handle bars, 150 wheels, 250 pedals, and more with flashcards,,! Used up first is known as the limiting reactant to correctly identify the limiting reactant Choice! Available/ # moles available/ # moles available/ # moles required comes from the balanced equation reactants products. 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Divide for each reactant: # moles required are left over study.! 3 moles of what substance is left over free response 2a ( part 1 of 2 limiting. The ER coefficient of Fe 2 O of Fe is number a 100 handle bars, 150 wheels, pedals! 0.10 mole, 100 mL of 1 M HCl is 0.10 mole, 100 mL of 1 M is. Coefficients in a chemical equation represent the a. masses, in grams of. Grams, of all reactants and products the one alternative that best completes the statement or answers the question and... Masses, in grams, of all reactants and products multiple-choice questions may continue on the column... Answers the question reagent in a reaction the smallest molar mass ( formula weight ) excess '' form ammonia to! That you might encounter in the kitchen a method to determine the percent yield of the reaction over! Reactant C ) catalyst D ) solid E ) gas 20 ) a catalyst is.! Up before the others to 1 mole ratio between reactants and products in Chemistry 1 edited NO + H O! 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