Question

when 70.0 grams of mno2 reacted with 128.0 grams of hcl, the reaction resulted in a 62.7% yield of chlorine gas. what is the actual yield of chlorine gas in grams? Mno2 + HCI —> MnCl2 + h2o + cl2​

Answer 1

35.8g of Cl₂ is the yield

Step-by-step 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) =

35.8g of Cl₂ is the yield