Question

MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O
If 39.1 grams of MnO2 reacts with 48.2 g of HCl, and 19.8 grams of Cl2 was collected, what is the percent yield?

Answer 1

First the limiting reactant should be determined:

$n(MnO_2)=\frac{39.1g}{86.94g/mol}=0.450 mol$

$n(HCl)=\frac{48.2g}{36.46g/mol}=1.32 mol$

$\frac{1.32}{0.450}=2.93<4$ , hence HCl is the limiting reactant (1 mole of MnO₂ requires 4 moles of HCl for the reaction). Therefore, theoretical yield of Cl₂ will be:

$m(Cl_2)=1.32molHCl\times\frac{1molCl_2}{4molHCl}\times\frac{70.9gCl_2}{1molCl_2}=23.4g$

$\%yield=\frac{19.8g}{23.4g}\times100=84.6\%$

Answer: 84.6%