Chemistry Chapter 12 Review. Remember to bring with you your Chapter 12 foldable, periodic table and a calculator if you wish to use one on the test. What is the amount of product formed when a reaction is carried out in the laboratory called?. Limiting reagent Actual yield Percent yield
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Remember to bring with you your Chapter 12 foldable, periodic table and a calculator if you wish to use one on the test.
2 CO + O2 2 CO2
5.4 mol CO (1 mol O2/2 mol CO) =
2.7 mol O2
3.6 L O2(1mol/22.4 L)(2mol CO/ 1 mol O2)(22.4 L/1 mol)
7.2 L CO
11.5 g CO ( 1 mol/28.01 g CO)(2 mol CO2/2 mol CO)(44.01 g CO2/1 mol)
18.07 g CO2
6.2 g O2 (1 mol/32.0 g) = .19 mol O2
4.0 g CO (1 mol/28.01) = .14 mol CO
.19 mol O2 (2 mol CO/1 mol O2) = .38 CO
.14 < .38
You don’t have enough CO so CO is the limiting reagent.
4.0 g CO(1 mol/28.01)(2 mol CO2/2 mol CO)(44.01 g CO2/1 mol) =
6.28 g CO2
(5.8 g/6.28 g) x 100 =
3 Zn + 2 MoO3 Mo2O3 + 3ZnO
5.4 molMoO3 (3 molZn/2 molMoO3) =
3.6 L MoO3(1mol/22.4 L)(3mol Zn/ 2 molMoO3)(22.4 L/1 mol)
11.5 g Zn ( 1 mol/65.39 g Zn)(3molZnO/3 molZn)(81.39 g ZnO/1 mol)
14.31 g ZnO
6.2 g Zn (1 mol/65.39 g) = .09 molZn
4.0 g MoO3(1 mol/143.94) = .03molMoO3
.09 molZn (2 molMoO3/3 molZn) = .06CO
You don’t have enough MoO3so MoO3is the limiting reagent.
4.0 g MoO3(1 mol/143.94)(3 molZnO/2 molMoO3)(81.39 g ZnO/1 mol) =
3.39 g ZnO
3 Zn + 2 MoO3 Mo2O3 + 3ZnOIf you actually did this reaction in a lab and produced 2.82 g ZnOwhat would your percent yield be?
(2.82 g/3.39 g) x 100 =