To determine the volume of ammonia produced when 10 g of hydrogen combines with nitrogen, we need to consider the balanced chemical equation for the reaction and apply the ideal gas law at standard temperature and pressure (STP).
The balanced chemical equation for the reaction between hydrogen (H2) and nitrogen (N2) to form ammonia (NH3) is:
3H2 + N2 → 2NH3
From the balanced equation, we can see that 3 moles of hydrogen react with 1 mole of nitrogen to produce 2 moles of ammonia.
First, let's calculate the number of moles of hydrogen (H2) present in 10 g using its molar mass:
Molar mass of H2 = 2 g/mol
Number of moles of H2 = Mass of H2 / Molar mass of H2 = 10 g / 2 g/mol = 5 mol
According to the stoichiometry of the balanced equation, 3 moles of hydrogen react to produce 2 moles of ammonia. Therefore, the number of moles of ammonia (NH3) produced can be calculated as:
Number of moles of NH3 = (Number of moles of H2 / 3) * 2 = (5 mol / 3) * 2 = 10/3 mol
At STP (standard temperature and pressure), 1 mole of any gas occupies a volume of 22.4 liters. Therefore, the volume of ammonia produced can be calculated as:
Volume of NH3 = Number of moles of NH3 * 22.4 L/mol = (10/3 mol) * 22.4 L/mol = 74.67 L
So, approximately 74.67 liters of ammonia at STP would be produced when 10 g of hydrogen reacts with nitrogen.