Question

1. Rusting involves the reaction of iron with oxygen to form iron(III) oxide, Fe 2 O 3 :

4 Fe (s) + 3 O 2 (g)  2 Fe 2 O 3 (s)
(a) What are the mole ratios of the product to each reactant and of the reactants to each
other?

(b) How many moles of iron(III) oxide are formed by the complete oxidation of 5.2 mol of
iron?

2. In the atmosphere, nitrogen dioxide reacts with water to produce NO and nitric acid, which
contributes to pollution by acid rain.
3 NO 2 (g) + H 2 O (l)  2 HNO 3 (aq) + NO (g)
How many grams of HNO 3 are produced for every 1.0 mole of NO 2 that reacts? The molecular
mass of HNO 3 is 63.0 amu.

Answer 1

1.

a)

4 mol Fe : 3 mol O2

3 mol O2 : 2 mol Fe2O3

2 mol Fe : 1 mol Fe2O3

b) 2.6 mol Fe2O3

2. 42 g HNO3

Step-by-step explanation:

1. a) 4 Fe (s) + 3 O 2 (g) -----> 2 Fe 2 O 3 (s)

from reaction 4 mol 3 mol 2 mol

4 mol Fe : 3 mol O2

3 mol O2 : 2 mol Fe2O3

4 mol Fe : 2 mol Fe2O3 = 2 mol Fe : 1 mol Fe2O3

b)

from reaction

2 mol Fe : 1 mol Fe2O3

5.2 mol Fe : x mol Fe2O3

x = 5.2 *1/2 = 2.6 mol Fe2O3

2.

3 NO 2 (g) + H 2 O (l)-----> 2 HNO 3 (aq) + NO (g)

from reaction 3 mol 2 mol

given 1 mol x mol

3 mol NO2 : 2 mol HNO3 = 1 mol NO2 : x mol HNO3

x = 1*2/3 = 2/3 mol HNO3

Molecular mass of HNO3 = 63.0 amu,

so molar mass of HNO3 = 63.0 g/mol.

2/3 mol HNO3 * 63.0 g/mol HNO3 = 42 g HNO3