Not all stoichiometry problems have to be limiting reagent problems. If the question defines that one of the reactant is in excess, you are being told that the limiting reagent is the other reactant. Use this to solve the first three warmup problems.
N2(g) + 3H2(g) → 2NH3(g)
When 3.00 moles of N2 react with excess hydrogen, how many moles of NH3 are formed?
Using the following balanced chemical equation, solve for the total number of moles in the final reaction mixture for each set of starting amounts for the reagents. The reaction is run at 273.15 K and 1 atm pressure.
H2(g) + Cl2(g) → 2HCl(g)
4.00 moles of H2 and 8.00 moles Cl2 The reactant react in a 1:1 ratio, therefore it is obvious that H2 is the limiting reactant and that only 4 moles of Cl2 will react with 4 moles in excess (or leftover) now we calculate the number of moles of product:
\[\rm 4\,mol\,H_2 \left(2\,mol\,HCl\over 1\,mol\,H_2\right) = 8\,mol\,HCl\]
and finally add up ALL the moles remaining: 8 mol HCl + 4 mol Cl2 = 12 mol TOTAL in the reaction mixture after the reaction is complete
6.00 moles H2 and 4.00 moles Cl2 now the Cl2 is the limiting reactant and that only 4 moles of H2 will react with 2 moles in excess (or leftover) now we calculate the number of moles of product:
\[\rm 4\,mol\,Cl_2 \left(2\,mol\,HCl\over 1\,mol\,Cl_2\right) = 8\,mol\,HCl\]
and finally add up ALL the moles remaining: 8 mol HCl + 2 mol H2 = 10 mol TOTAL in the reaction mixture after the reaction is complete
12.0 L H2 and 20.0 L Cl2 Because of Avogadro's Law, we can work this problem in volume units (liters) and then convert to moles at the end. This means that the H2 is the limiting reactant (12 L is less than 20 L) and that a matching 12 L of Cl2 will react with it and 8 L will be in excess (or leftover). Now we calculate the number of liters of product:
\[\rm 12\,L\,H_2 \left(2\,L\,HCl\over 1\,L\,H_2\right) = 24\,L\,HCl\]
and finally add up ALL the volumes remaining: 24 L HCl + 8 L Cl2 = 32 L TOTAL for the reaction mixture after the reaction is complete. To convert to moles you can use standard molar volume at STP which happens to be 22.4 L/mol, therefore:
Solve the next three questions using the following balanced chemical equation:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
12.6 L of C3H8 and 25.2 L of O2 react to completion. What is the final volume of all gases, assuming constant temperature and pressure? To get the limiting reactant let's calculate how many liters of reaction we can run with each amount given.
\[\rm 12.6\,L\,C_3H_8 \left(1\,L\,rxn\over 1\,L\,C_3H_8\right) = 12.6\,L\,of\,rxn\]
\[\rm 25.2\,L\,O_2 \left(1\,L\,rxn\over 5\,L\,O_2\right) = 5.04\,L\,of\,rxn\]
So we now know that the oxygen is the limiting reactant and will run the reaction as written a total of 5.04 times. Using this we can get the scaled volumes of the products:
\[\rm 5.04\,L\,rxn \left(3\,L\,CO_2\over 1\,L\,rxn\right) = 15.12\,L\,of\,CO_2\]
\[\rm 5.04\,L\,rxn \left(4\,L\,H_2O\over 1\,L\,rxn\right) = 20.16\,L\,of\,H_2O\]
as for leftovers (the amount of excess C3H8) we know that only 5.04 L of rxn were run from the 12.6 L of rxn possible which means that 7.56 L of rxn remain and because the rxn to C3H8 ratio is 1:1, then 7.56 L of C3H8 will remain. So to sum up all species:
15.12 L CO2 + 20.16 L of H2O + 7.56 L of C3H8 = 42.84 L total volume of all species after the reaction.
1.35 moles of C3H8 and 8.51 moles O2 react to completion. What is the final volume if all the gases in the final reaction mixture are carefully stored at 400 K and 2.54 atm? calculate moles of reaction from each reactant:
\[\rm 1.35\,mol\,C_3H_8 \left(1\,mol\,rxn\over 1\,mol\,C_3H_8\right) = 1.35\,L\,of\,rxn\]
\[\rm 8.51\,mol\,O_2 \left(1\,mol\,rxn\over 5\,mol\,O_2\right) = 1.702\,mol\,of\,rxn\]
So we now know that the C3H8 is the limiting reactant and will run the reaction as written a total of 1.35 times. Using this we can get the scaled number of moles of the products:
\[\rm 1.35\,mol\,rxn \left(3\,mol\,CO_2\over 1\,mol\,rxn\right) = 4.05\,mol\,of\,CO_2\]
\[\rm 1.35\,mol\,rxn \left(4\,mol\,H_2O\over 1\,mol\,rxn\right) = 5.40\,mol\,of\,H_2O\]
as for leftovers (the amount of excess O2) we know that only 1.35 mol of rxn were run from the 1.702 mol of rxn possible which means that 0.352 mol of rxn remain in oxygen moles. The oxygen is in a 5:1 ratio to the whole reaction which means that 0.352 mol rxn is equal to 1.76 mol O2. So to sum up all species in the mixture in moles:
4.05 mol CO2 + 5.40 mol of H2O + 1.76 mol of O2 = 11.21 mol total moles of all species after the reaction. Now use the ideal gas law to convert that to volume:
\(\displaystyle V = {nRT\over P} = {11.21(0.08206)400\over 2.54}\) = 144.9 L total
This same reaction is run at a much colder temperature to ensure that the water produced by the reaction is in the liquid phase. Now you react 8.55 L of C3H8 with 18.5 L O2. What is the total volume of all species in the final reaction mixture? using the same logic as in previous problems using volumes we find you can run 8.55 L of rxn with the C3H8 and only 3.70 L rxn with the O2. So the oxygen is the limiting reactant and 4.85 L of C3H8 will be in excess. For gas volume, we only need the CO2 and not the water (its a liquid).
\[\rm 3.70\,L\,rxn \left(3\,L\,CO_2\over 1\,L\,rxn\right) = 11.10\,L\,of\,CO_2\]
Now we sum up all gas species in the mixture in liters:
11.10 L CO2 + 4.85 L of C3H8 = 15.95 L total gases after the reaction.
S22