Multiple choice

60ml of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38ml of $N_{2}$ was formed, calculate the volume of each gas in the mixture.

  1. $\displaystyle \:NO= 44ml; N_{2}O= 16ml$
  2. $\displaystyle \:NO= 45ml; N_{2}O= 20ml$
  3. $\displaystyle \:NO= 34ml; N_{2}O= 22ml$
  4. $\displaystyle \:NO= 20ml; N_{2}O= 26ml$
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A Correct answer
Explanation

Reactions: N2O + H2 -> N2 + H2O; 2NO + 2H2 -> N2 + 2H2O. Let x be vol of N2O and y be vol of NO. x + y = 60. From stoichiometry, x + y/2 = 38. Solving: y/2 = 22, y = 44. Then x = 16.

AI explanation

Let the volume of nitrous oxide (N2O) be x ml and the volume of nitric oxide (NO) be (60 - x) ml. The reactions with hydrogen are N2O + H2 -> N2 + H2O and 2NO + 2H2 -> N2 + 2H2O. Using Gay-Lussac's law of combining volumes, x ml of N2O produces x ml of N2, and (60 - x) ml of NO produces (60 - x)/2 ml of N2. The total volume of N2 formed is x + (60 - x)/2 = 38. Solving this equation yields x = 16, meaning the mixture contains 16 ml of N2O and 44 ml of NO.

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