Answer:
35.8g of Cl₂ is the yield
Explanation:
Based on the reaction:
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
1 mole of MnO₂ and 4 moles of HCl react producing 1 mole of Cl₂
To solve this question we must find limiting reactant. with limiting reactant we can find the theoretical yield of Cl₂. As the actual yield is the 62.7% we can find actual yield of Cl₂ in grams:
Moles MnO₂ -Molar mass: 86.9368g/mol-:
70.0g * (1mol / 86.9368g) = 0.805 moles
Moles HCl -Molar mass: 36.46g/mol-:
128.0g * (1mol / 36.46g) = 3.51 moles
For a complete reaction of 3.51 moles of HCl are required:
3.51 moles HCl * (1mol MnO₂ / 4mol HCl) = 0.878 moles MnO₂.
As there are just 0.805 moles of MnO₂, MnO₂ is limiting reactant.
1 mole of MnO₂ produce 1 mole of Cl₂. The theoretical moles of Cl₂ produced are 0.805 moles.
As the yield is of 62.7%, the yield of Cl₂ is:
0.805 moles * (62.7 / 100) = 0.505 moles Cl₂. In grams:
0.505 moles Cl₂ * (70.906g / mol) =
