What mass of ammonia is produced when 1.73 L of nitrogen (STP)react completely in the following equation?
N2 + H3 > 2NH3
1.73L N2 x (2 mol NH3/1 mol N2) = ? mol NH3.
g NH3 = mol NH3 x molar mass NH3.
To determine the mass of ammonia produced, we need to use the given volume of nitrogen and the stoichiometry of the balanced chemical equation.
First, let's find the number of moles of nitrogen. We know that the volume of nitrogen is given at standard temperature and pressure (STP), which is 1.73 L. At STP, 1 mole of any gas occupies 22.4 L. So, we can use the following conversion:
1.73 L N2 x (1 mole N2 / 22.4 L N2) = 0.0772 moles N2
Now, let's use the stoichiometry of the balanced chemical equation to determine the number of moles of ammonia produced. From the equation, we see that 1 mole of nitrogen reacts to produce 2 moles of ammonia. Therefore, the number of moles of ammonia produced can be calculated as follows:
0.0772 moles N2 x (2 moles NH3 / 1 mole N2) = 0.154 moles NH3
Finally, we can convert the moles of ammonia to grams using its molar mass. The molar mass of ammonia (NH3) is 17.03 g/mol. Therefore, the mass of ammonia can be calculated as follows:
0.154 moles NH3 x (17.03 g / 1 mole NH3) = 2.63 g NH3
Therefore, approximately 2.63 grams of ammonia will be produced when 1.73 L of nitrogen reacts completely in the given equation.