Question #150625

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?

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?

Expert's answer

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 MnO_{2} requires 4 moles of HCl for the reaction). Therefore, theoretical yield of Cl_{2} 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%

Learn more about our help with Assignments: Inorganic Chemistry

## Comments

## Leave a comment