In this question, we have the following reaction:
4 HCl + MnO2 -> MnCl2 + 2 H2O + Cl2
The reaction is already properly balanced, therefore we can take the molar ratios from it:
4 HCl = 1 MnO2
4 HCl = 1 Cl2
1 MnO2 = 1 Cl2
This shows us the molar ratio between the compounds, which means the number of moles needed for each compound in order for the reaction to occur, and this will also help us find the limiting and excess reactant
Now let's take the mass of each compound given in the question:
40.3 grams of MnO2, molar mass 86.94g/mol
45.7 grams of HCl, molar mass 36.46g/mol
Now let's check which one is the limiting, let's start with MnO2:
86.94g = 1 mol
40.3g = x moles
x = 0.463 moles of MnO2 in 40.3 grams
We know that whatever number of moles of MnO2 we have, we will require 4 times that of HCl, therefore
0.463 * 4 = 1.85 moles of HCl are required, but we don't know how much HCl we have, let's calculate it:
36.46g = 1 mol
45.7g = x moles
x = 1.25 moles, therefore we have less HCl than we actually need, this makes HCl the limiting reactant, and we have an excess of MnO2
Now for the theoretical yield of Cl2, we will use the number of moles of the limiting reactant, HCl 1.25 moles, in order to find the mass of Cl2, and again we need to look at the molar ratio, which is also 4:1, so the number of moles of Cl is HCl/4
Cl2 = HCl/4
Cl2 = 1.25/4
Cl2 = 0.312 moles
Now we take the molar mass of Cl2, 70.91g/mol, and find the final mass
70.91g = 1 mol
x grams = 0.312 moles
x = 22.12 grams of Cl2 is the theoretical yield
If we only have 85.1% of actual yield, the mass will be:
22.12g = 100%
x grams = 85.1%
x = 18.82 grams
Limiting reactant = HCl
Theoretical yield = 22.12 grams
Actual yield 85.1% = 18.82 grams