What mass of MnCl2 is produced when 0.091 g of Cl2 is generated?

MMn = 54.94g/mol
MO2 = 2(16) = 32g/mol
MH = 1g/mol
MCl = 35.45g/mol
Molar Mass of MnO2:
54.94
+
2
(
16
)
=
86.94
Molar Mass of HCl:
1
+
35.45
=
36.45
Looking at the equation, we need a 4:1 ratio of HCl to MnO2 to make the reaction work, let's see if we have four times as much of the HCl.
Mols of MnO2:
42.7
86.94
=
0.49
Mols of HCl:
47.1
36.45
=
1.29
As we can see, we have less than the amount of HCl we need for all of the MnO2 to react, therefore the HCl is the limiting reactant.
For the theoretical yield of Cl2, we need to determine how many mols will be produced. As we can see by the equation, four mols of HCl make one mol of Cl2.
Molar Mass of Cl2:
35.45
⋅
2
=
70.9
Since we have 1.29mol of HCl, and we know it is the limiting reactant:
Mols of Cl2
1.29
4
=
0.323
Mass of Cl2 (Theoretical yield)
0.323
⋅
70.9
=
22.9
To calculate the actual yield, we multiply the theoretical yield by the final percentage:
22.9
⋅
0.791
=
18.11
Therefore, the actual yield of chlorine is 18.11

A.