Assuming that pressure and temperature are constant, how many liters of oxygen gas will be required to produce 0.160 L of nitrogen gas. 4 NH3 (g) + 3 O2 (g) → 6 H2O (g) + 2 N2 (g)

0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.

Explanation:

From the reaction, 3 moles of oxygen react with ammonia to produce 2 moles of nitrogen gas

So, 3 moles of 02 = 2 moles of N2

Since I mole of a gas occupies 22.4dm^3 or 22,4 L of the gas

3 * 22,4L of O2 produces 2 * 22.4 L of N2

67.2 L of O2 produces 44.8 L of N2

(67.2 * 0.160 / 44.8) L of O2 will produce 0.160L of N2

10.752 / 44,8 L = 0.24 L of O2 will produce 0.160 L of N2

0.24 L of oxygen gas will be required to produce 0.160 L of nitrogen gas.