If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the reaction proceeds to 82.0% yield?

2N2 (g) + O2 (g) = 2N2 O (g)

Question

Carla

Chemistry

20 October, 06:32

If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the reaction proceeds to 82.0% yield?

2N2 (g) + O2 (g) = 2N2 O (g)

+1

Answers (1)

Know the Answer?

Olive Dorsey · Answer 1

259 mL Assuming the the nitrogen and oxygen gasses are at the same pressures, then the ratios of the moles of the gasses will be in proportion to their volumes according to the ideal gas law. So we have 316 / 178 = 1.775 moles of nitrogen gas per mole of oxygen gas. This means that the nitrogen gas will be the limiting reactant. For each mole of nitrogen gas used, we will get 1 mole of N2O. So the resulting volume of N2O with 100% yield will be the same as the volume of nitrogen gas used, so 100% yield will produce 316 mL. But since we only have 82.0% yield, that means that the volume of N2O produced will be 316 mL * 0.820 = 259.12 mL. Rounding to 3 significant figures, gives 259 mL. So assuming constant pressure and temperature, the volume of N2O will be 259 mL.

If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the reaction proceeds to 82.0% yield?2N2 (g) + O2 (g) = 2N2 O (g)

If 316 mL nitrogen is combined with 178 mL oxygen, what volume of N2O is produced at constant temperature and pressure if the reaction proceeds to 82.0% yield?

2N2 (g) + O2 (g) = 2N2 O (g)