Assume that you have 1.37 of H2 and 3.42 of N2 . How many grams of ammonia can y
ID: 752954 • Letter: A
Question
Assume that you have 1.37 of H2 and 3.42 of N2 . How many grams of ammonia can you make, and how many grams of which reactant will be left over? 3H2+N2->2NH3Explanation / Answer
The reaction calls for 3 molecules of H2 and 1 molecule of N2. We have 1.44 mol of H2 and 3.42 mol of N2. We see how many molecules of NH3 we would be able to make if there were to be an excess of all other molecules. H2 --> 1.44mol H2 * 2 mol NH3 / 3 mol H2 = .96 mol NH3 N2 --> 3.42mol N2 * 2 mol NH3 / 1 mol N2 = 6.84 mol NH3 Since H2 runs out before N2, we can see that N2 will be the leftover. This is called the excess reactant. The reactant that runs out first, in this case H2, is called the limiting reactant (sometimes called the limiting reagent). We can see that .96 mol of NH3 will be made. We then find the molar mass of that compound: N - 14.00674g/mol H - 1.00794g/mol The molar mass of NH3 = (14.00674 + 1.00794 * 3)g/mol = 17.03056g/mol. So the mass of ammonia is .96mol * 17.03056g/mol = 16.35g. Then we find how many moles of N2 is left over. This can be done by finding how many moles of N2 was actually used to make .96mol of NH3. .96mol NH3 * 1 mol N2 / 2 mol NH3 = .48mol N2 Since we started off with 3.42mol N2, we are left with 3.42mol - .48mol = 2.94mol N2. Now we find the mass of it, first finding the molar mass of N2 and multiplying the # of moles with the molar mass. Molar mass of N2 = (14.00674 * 2)g/mol = 28.01348g/mol 2.94mol N2 * 28.01348g/mol = 82.36g