Mole to Mass Conversion Problems Answer Key (CHEM 1405)

Back to Chemistry 1405 Practice Problems

Balance the equations, then solve for the required substance using dimensional analysis.

1)  How many moles of oxygen gas are required to produce 100.0 grams of water?

            2H₂   +    O₂             2H₂O

$100.0gH_2O\times(\frac{1molH_2O}{18.02gH_2O})\times(\frac{1molO_2}{2molH_2O})=$ 2.775 mol O₂

 

 

 

2)  Calculate the number of moles of hydrochloric acid required to react with 2.57 grams of calcium hydroxide.

            2HCl       +        Ca(OH)₂           CaCl₂    +     2H₂O

$2.57gCa(OH)_2\times(\frac{1molCa(OH)_2}{74.10gCa(OH)_2})\times(\frac{2molHCl}{1molCa(OH)_2})=$ 0.0694 mol HCl

 

b.  How many grams of calcium hydroxide are needed to produce 4.45 moles of water?

$4.45molH_2O\times(\frac{1molCa(OH)_2}{2molH_2O})\times(\frac{74.10gCa(OH)_2}{1molCa(OH)_2})=$ 165 g Ca(OH)₂

 

 

 

3)  How many moles of aluminum will be produced from 30.0 kg of aluminum oxide?

            2Al₂O₃          4Al     +     3O₂ 

$30.0kgAl_2O_3\times(\frac{1000gAl_2O_3}{1kgAl_2O_3})\times(\frac{1molAl_2O_3}{101.96gAl_2O_3})\times(\frac{4molAl}{2molAl_2O_3})=$ 588 mol Al

 

b.  If 52 g of aluminum are produced, how many moles of aluminum oxide are needed?

$52gAl\times(\frac{1molAl}{26.98gAl})\times(\frac{2molAl_2O_3}{4molAl})=$ 0.96 mol Al₂O₃ 

 

 

 

4)  Calculate the number of moles of Na₂CO₃ required to make 100.0 g of NaNO₃.

            Na₂CO₃    +      2HNO₃           2NaNO₃    +     CO₂    +     H₂O

$100.0gNaNO_3\times(\frac{1molNaNO_3}{85.00gNaNO_3})\times(\frac{1molNa_2CO_3}{2molNaNO_3})=$ 0.5882 mol Na₂CO₃

 

b.  If 7.50 g of Na₂CO₃ reacts, how many moles of CO₂ are produced?

$7.50gNa_2CO_3\times(\frac{1molNa_2CO_3}{105.99gNa_2CO_3})\times(\frac{1molCO_2}{1molNa_2CO_3})=$ 0.0708 mol CO₂

 

 

 

5)  If 625 g of Fe₂O₃ are produced in the reaction, how many moles of hydrogen gas are produced at the same time?

            2Fe      +     3H₂O            3H₂     +     Fe₂O₃

$625gFe_2O_3\times(\frac{1molFe_2O_3}{159.7gFe_2O_3})\times(\frac{3molH_2}{1molFe_2O_3})=$ 11.7 mol H₂

 

b.  How many moles of iron would be needed to generate 27 g of hydrogen gas?

$27gH_2\times(\frac{1molH_2}{2.02gH_2})\times(\frac{2molFe}{3molH_2})=$ 8.9 mol Fe

 

c.  If 52 moles of water are used, how many grams of iron (III) oxide would be produced?

$52molH_2O\times(\frac{1molFe_2O_3}{3molH_2})\times(\frac{159.7gFe_2O_3}{1molFe_2O_3})=$ 2800 g Fe₂O₃

 

6)  If 0.905 moles of Al₂O₃ is produced in the following reaction, what mass of iron is also produced?

            Fe₂O₃      +      2Al            2Fe     +      Al₂O₃

$0.905molAl_2O_3\times(\frac{2molFe}{1molAl_2O_3})\times(\frac{55.85gFe}{1molFe})=$ 101 g Fe

 

b.  How many moles of Fe₂O₃ will react with 99.0 g of aluminum?

$99.0gAl\times(\frac{1molAl}{26.98gAl})\times(\frac{1molFe_2O_3}{2molAl})=$ 1.83 mol Fe₂O₃

7)  If 10.00 g of ammonia (NH₃) react, how many moles of ammonium hydrogen phosphate will be produced?

            H₃PO₄     +     2NH₃          (NH₄)₂HPO₄

$10.00gNH_3\times(\frac{1molNH_3}{17.04gNH_3})\times(\frac{1mol(NH_4)_2HPO_4}{2molNH_3})=$ 0.2934 mol (NH₄)₂HPO₄

 

b.  If 280 grams of product are produced, how many moles of phosphoric acid are required?

$280g(NH_4)_2HPO_4\times(\frac{1mol(NH_4)_2HPO_4}{132.06g(NH_4)_2HPO_4})\times(\frac{1molH_3PO_4}{1mol(NH_4)_2HPO_4})=$ 2.1 mol H₃PO₄

 

 

 

8)  How many moles of nitrogen gas are produced when 36.0 g of NH₄NO₃ reacts?

            2NH₄NO₃            2N₂    +      O₂      +      4H₂O

$36.0gNH_4NO_3\times(\frac{1molNH_4NO_3}{80.06gNH_4NO_3})\times(\frac{2molN_2}{2molNH_4NO_3})=$ 0.450 mol N₂

 

b.  If 7.35 moles of water are produced, what was the mass of NH₄NO₃ that reacted?

$7.35molH_2O\times(\frac{2molNH_4NO_3}{4molH_2O})\times(\frac{80.06gNH_4NO_3}{1molNH_4NO_3})=$ 294 g NH₄NO₃