Question

If 0.86 mole of MnO₂ and 56.6 g of HCl react, how many grams of Cl₂ will be produced?

MnO₂ + 4HCI —> MnCl₂ + Cl₂ + 2H₂O
Be sure your answer has the correct number of significant digits.

Answer 1

To find the grams of Cl₂ produced, you can use the given information and the stoichiometry of the reaction.

First, calculate the moles of HCl:

Given mass of HCl = 56.6 g

Molar mass of HCl = 1 g/mol (for hydrogen) + 35.5 g/mol (for chlorine) = 36.5 g/mol

Moles of HCl = (Given mass) / (Molar mass) = 56.6 g / 36.5 g/mol ≈ 1.5493 moles

Now, according to the balanced chemical equation, 1 mole of MnO₂ reacts with 4 moles of HCl to produce 1 mole of Cl₂. Therefore, we can say that the moles of Cl₂ produced will be the same as the moles of HCl consumed.

So, moles of Cl₂ produced ≈ 1.5493 moles

Now, calculate the molar mass of Cl₂:

Molar mass of Cl₂ = 2 (atomic mass of Cl) = 2 * 35.5 g/mol = 71 g/mol

Finally, calculate the grams of Cl₂ produced:

Grams of Cl₂ = (Moles of Cl₂) * (Molar mass of Cl₂) = 1.5493 moles * 71 g/mol ≈ 109.857 g

Rounded to the correct number of significant digits, the answer is approximately 110 g of Cl₂ will be produced.

Step-by-step explanation: