Answer:
a. HCl
b. 24.3 g of Cl₂
c. 19.4 g of Cl₂
Explanation:
a. In order to determine the limiting reactant we convert the mass of each reactant to moles.
The reaction is: 4HCl (aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O (l) + Cl₂(g)
37.3 g / 86.94g/mol = 0.429 moles of manganese dioxide
50.1 g / 36.45 g/mol = 1.37 moles of HCl
Ratio is 1:4. 1 mol of MnO₂ needs 4 moles of hydrochloric to react
Then, 0.429 moles of dioxide will react with (0.429 . 4) /1 = 1.71 moles of HCl.
At least, we have 1.37 moles, so the HCl is the limiting reactant.
b. In this question, we can work with stoichiometry with a rule of three:
4 moles of HCl produce 1 mol of chlorine
Then, 1.37 moles of HCl will produce (1.37 . 1) /4 = 0.3425 moles of Cl₂
We convert the moles to mass → 24.3 g of Cl₂
c. In order to determine the actual yield of chlorine, we multiply
Theoretical yield . Yield percent/100 → 24.3 g . 79.9/100 = 19.4 g
