ideal gas law and stoichiometry

Question

what volume of O2 at 760mmHg and 35C is required to synthesize 14.0mol of NO? balanced eqn= 4NH3+5O2-->4NO+6H2O

Answer 1

First use the mole ratio from the balanced equation:

4NH3 + 5O2 → 4NO + 6H2O

For every 4 mol of NO, you need 5 mol of O2.

14.0 mol NO × (5 mol O2 / 4 mol NO) = 17.5 mol O2

Now use the ideal gas law.

760 mmHg = 1.00 atm

35°C = 308.15 K

Solve for volume:

V = nRT/P

V = (17.5)(0.08206)(308.15)/(1.00)

V ≈ 442 L

So about 442 L of O2 is required.