Stoichiometry Review Answer Key (CHEM 1405)

Back to Chemistry 1405 Practice Problems

Balance the equation and write the molar ratios as indicated.

1.                                               Sn   +     4HgNO₃          Sn(NO₃)₄   +      4Hg

 

Molar ratio of Sn to Hg:         _1 mol Sn: 4 mol Hg_____       Molar ratio of HgNO₃ to Hg:      _4 mol HgNO₃: 4 mol Hg__

Molar ratio of Sn to HgNO₃: _1 mol Sn: 4 mol HgNO₃_       Molar ration of Sn to Sn(NO₃)₄: _1 mol Sn: 1 mol Sn(NO₃)₂_

 

 

Balance each equation.  Show all your work.  Be sure to circle answers and check for significant figures.

2.  Aluminum bromide is reacted with chlorine to produce aluminum chloride and bromine.

2AlBr₃     +     3Cl₂          2AlCl₃     +     3Br₂

a) How many grams of chlorine would be produced from 35.4 grams of aluminum chloride?

$35.4gAlCl_3\times(\frac{1molAlCl_3}{133.33gAlCl_3})\times(\frac{3molCl_2}{2molAlCl_3})\times(\frac{70.90gCl_2}{1molCl_2})=$  28.2g Cl2

 

b) If you produce 452 grams of bromine, how many grams of aluminum bromide did you have originally?

$452gBr_2\times(\frac{1molBr_2}{159.8gBr_2})\times(\frac{2molAlBr_3}{3molBr_2})\times(\frac{266.68gAlBr_3}{1molAlBr_3})=$ 503g AlBr₃

 

 

 

3.  Hydrogen and nitrogen combine to yield hydrogen nitride.

3H₂     +     N₂          2H₃N

a) How many grams of nitrogen would be needed to produce 17.8 grams of hydrogen nitride?

$17.8gH_3N\times(\frac{1molH_3N}{17.04gH_3N})\times(\frac{1molN_2}{2molH_3N})\times(\frac{28.02gN_2}{1molN_2})=$  14.6g N₂

 

b) How many grams of hydrogen nitride are produced if you have 645 grams of nitrogen?

$645gN_2\times(\frac{1molN_2}{28.02gN_2})\times(\frac{2molH_3N}{1molN_2})\times(\frac{17.04gH_3N}{1molH_3N})=$  784g H₃N

 

 

 

4.  Zinc chlorate is reacted with sodium to yield sodium chlorate and zinc.

Zn(ClO₃)₂     +     2Na          Zn     +     2NaClO₃

a) How many grams of zinc would be produced from 4.56 grams of sodium?

$4.56gNa\times(\frac{1molNa}{22.99gNa})\times(\frac{1molZn}{2molNa})\times(\frac{65.39gZn}{1molZn})=$  6.48g Zn

 

b) How many grams of zinc would be produced from 21.45 grams of zinc chlorate?

$21.45gZn(ClO_3)_2\times(\frac{1molNa}{232.29gZn(ClO_3)_2})\times(\frac{1molZn}{1molZn(ClO_3)_2})\times(\frac{65.39gZn}{1molZn})=$

 6.038g Zn

 

c) Based on the previous answers, which reactant is the limiting reactant? __Zn(ClO₃)₂__

d) When this experiment was carried out in the lab, the actual yield of zinc was 5.45 grams. What is the percent yield?

$\%yield=\frac{5.45gZn}{6.038gZn}\times100\%=$ 90.3%

 

 

 

5.  Aspirin (C₉H₆O₃) and water are produced when salicylic acid (C₇H₆O₃) reacts with acetic anhydride (C₄H₆O₃).

2C₇H₆O₃     +     C₄H₆O₃          2C₉H₆O₃     +     3H₂O

a) If 47.5 grams of salicylic acid reacts with 34.4 g of acetic anhydride, what is the limiting reagent?

Limiting Reagent                                                                                                                    Theoretical Yield
$47.5gC_7H_6O_3\times(\frac{1molC_7H_6O_3}{138.13gC_7H_6O_3})\times(\frac{3molH_2O}{2molC_7H_6O_3})\times(\frac{18.02gH_20}{1molH_2O})=$ 9.30g H₂O

 

$34.4gC_4H_6O_3\times(\frac{1molC_4H_6O_3}{102.10gC_4H_6O_3})\times(\frac{3molH_2O}{1molC_4H_6O_3})\times(\frac{18.02gH_20}{1molH_2O})=$ 18.2g H₂O

Salicylic acid (C₇H₆O₃) is the limiting reagent.

 

b) How much of each product is produced in this reaction?

9.30g H₂O (work shown in part a)

$47.5gC_7H_6O_3\times(\frac{1molC_7H_6O_3}{138.13gC_7H_6O_3})\times(\frac{2molC_9H_6O_3}{2molC_7H_6O_3})\times(\frac{162.15gC_9H_6O_3}{1molC_9H_6O_3})=$ 55.8g C₉H₆O₃ 

 

c) How much excess reagent is left over after the reaction has taken place?

$47.5gC_7H_6O_3\times(\frac{1molC_7H_6O_3}{138.13gC_7H_6O_3})\times(\frac{1molC_4H_6O_3}{2molC_7H_6O_3})\times(\frac{102.10gC_4H_6O_3}{1molC_4H_6O_3})=17.6gC_4H_6O_3$ (used up)

34.4g C₄H₆O₃ (total) - 17.6g C₄H₆O₃ (used up) = 16.8g C₄H₆O₃ left over in excess

 

d) What is the percent yield if 57.4 g of aspirin is produced?

$\%yield=\frac{57.4gC_9H_6O_3}{55.8gC_9H_6O_3}\times100\%=$ 103%

 

 

 

6.  Elemental copper can be produced by reacting iron (II) with copper II sulfate solution.

Fe     +     CuSO₄          Cu     +     FeSO₄

This experiment was carried out, the resulting copper was filtered and dried and data collected:

     Mass of iron                                    4.20 grams

     Mass of filter paper                        0.67 grams

     Mass of filter paper and copper   4.51 grams

a) What is the experimental (actual) yield of copper in this experiment in grams?

     4.51g (filter paper and copper)
    -0.67g (filter paper)___________
     3.84g (actual yield of copper)

 

b) What is the theoretical yield of copper in this experiment (assuming iron is the limiting reagent)?

$4.20gFe\times(\frac{1molFe}{55.85gFe})\times(\frac{1molCu}{1molFe})\times(\frac{63.55gCu}{1molCu})=$ 4.78g Cu

 

c) What is the percent yield in this experiment?

$\%yield=\frac{3.84gCu}{4.78gCu}\times100\%=$ 80.3%