Answer:
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 MnO₂ reacts with 4 moles of HCl to produce 1 mol of Cl₂. So, we can write the following mole ratios:
1 mol MnO₂/4 mol HCl or 4 mol HCl/1 mol MnO₂
1 mol MnO₂/1 mol Cl₂
4 mol HCl/1 mol Cl₂
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(MnO₂) = 54.9 g/mol Mn + (16 g/mol x 2 O) = 86.9 g/mol
1 mol MnO₂ x 86.9 g/mol = 86.9 g MnO₂
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(Cl₂) = 2 x 35.4 g/mol Cl = 70.8 g/mol
1 mol Cl₂ x 70.8 g/mol = 70.8 g Cl₂
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 MnO₂. We multiply the actual amount of MnO₂ 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 MnO₂, and we have 128 grams of HCl. So, the reactant in excess is HCl, and <em>the limiting reactant is MnO₂</em>.
With the limiting reactant, we calculate the theoretical yield of Cl₂. We use the stoichiometric ratio 70.8 g Cl₂/86.9 g MnO₂:
70.0 g MnO₂ x 70.8 g Cl₂/86.9 g MnO₂ = 57 g Cl₂
Finally, we calculate the actual yield of chlorine gas (Cl₂), 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
