If 250.3 L of hydrogen gas d = 0.0899 g/L) reacts with an excess of nitrogen gas, what mass of ammonia would be produced?

Question

Quizplus · Accepted Answer

The mass of hydrogen gas can be calculated using its density and volume: mass = density × volume = 0.0899 g/L × 250.3 L = 22.5 g. The balanced chemical equation for the synthesis of ammonia (NH3) from nitrogen (N2) and hydrogen (H2) is: N2 + 3H2 → 2NH3. From the equation, 3 moles of H2 produce 2 moles of NH3. The molar mass of H2 is approximately 2 g/mol, so 22.5 g of H2 is 22.5 / 2 = 11.25 moles. This amount of hydrogen would produce (11.25 moles H2 × 2 moles NH3 / 3 moles H2) = 7.5 moles of NH3. The molar mass of NH3 is approximately 17 g/mol, so the mass of NH3 produced is 7.5 moles × 17 g/mol = 127.5 g, which rounds to 127 g.

If 2503 L of Hydrogen Gas D = 0