35.7 g
Explanation:
yield(%) = actual yield/theoretical yield x 100
We can calculate the theoretical yield by considering the balanced chemical equation:
MnO₂ + 4HCI —> MnCl₂ + 2H₂O + Cl₂
According to the equation, 1 mol of MnOâ‚‚ reacts with 4 moles of HCl to produce 1 mol of Clâ‚‚. So, we can write the following mole ratios:
1 mol MnOâ‚‚/4 mol HCl or 4 mol HCl/1 mol MnOâ‚‚
1 mol MnOâ‚‚/1 mol Clâ‚‚
4 mol HCl/1 mol Clâ‚‚
As we have the amounts of reactants in grams, we have to convert from moles to mass by using the molecular weight (MW) of each compound:
MW(MnOâ‚‚) = 54.9 g/mol Mn + (16 g/mol x 2 O) = 86.9 g/mol
1 mol MnOâ‚‚ x 86.9 g/mol = 86.9 g MnOâ‚‚
MW(HCl) = 1 g/mol H + 35.4 g/mol Cl = 36.4 g/mol
4 moles HCl x 36.4 g/mol = 145.6 g HCl
MW(Clâ‚‚) = 2 x 35.4 g/mol Cl = 70.8 g/mol
1 mol Clâ‚‚ x 70.8 g/mol = 70.8 g Clâ‚‚
Now, we have to figure out which is the limiting reactant. For this, we use the stoichiometric ratio: 145.6 g HCl/86.9 g MnOâ‚‚. We multiply the actual amount of MnOâ‚‚ by the stoichiometric ratio:
70.0 g MnO₂ x 145.6 g HCl/86.9 g MnO₂ = 117.3 g HCl  < 128.0 g HCl
We need 117.3 grams of HCl to completely react with 70.0 grams of MnOâ‚‚, and we have 128 grams of HCl. So, the reactant in excess is HCl, and the limiting reactant is MnOâ‚‚.
With the limiting reactant, we calculate the theoretical yield of Clâ‚‚. We use the stoichiometric ratio 70.8 g Clâ‚‚/86.9 g MnOâ‚‚:
70.0 g MnOâ‚‚ x 70.8 g Clâ‚‚/86.9 g MnOâ‚‚ = 57 g Clâ‚‚
Finally, we calculate the actual yield of chlorine gas (Clâ‚‚), by using the first equation:
yield(%) = actual yield/theoretical yield x 100
⇒ actual yield = theoretical yield x yield(%)/100
            = 57 g x 62.7%/100
            = 35.7 g