Question

How many grams of oxygen gas would have been required to produce 1000 grams of nitric acid HNO3?

Answer 1

4NH₃ + 5O₂ → 4NO + 6H₂O (1)

2NO + O₂ → 2NO₂ (2)

3NO₂ + H₂O → 2HNO₃ + NO (3)

M(HNO₃) = 63.01 g/mol

n(HNO₃) $= \frac{m}{M} = \frac{1000}{63.01} = 15.87 \;mol$

According to the reaction (3):

n(NO₂) = $\frac{3}{2}$ n(HNO₃) $= \frac{3}{2} \times 15.87 = 23.80 \;mol$

According to the reaction (2):

n(NO) = n(NO₂) = 23.80 mol

According to the reaction (1):

n(O₂) = $\frac{5}{4}$ n(NO) $= \frac{5}{4} \times 23.80 = 29.75 \;mol$

M(O₂) = 15.99 g/mol

m(O₂) $= 15.99 \times 29.75 = 475.70 \;g$

Answer: 475.70 g