Question

Chlorine gas can be produced by the reaction of hydrochloric acid with manganese(IV) oxide. If the percent yield for the reaction is 91%, how many grams of chlorine will be produced by the reaction of 212 g of HCl with excess MnO2?

Answer 1

$MnO_2(s) + 4HCl(aq) → Cl_2(g) + MnCl_2(aq) + 2H_2O(l)$

from the above equation we can see that for every 4 mole of HCl consumed 1 mole of chlorine gas is produced.

total moles of HCl =$\frac{212}{36.5}$=5.8

moles of chlorine gas produced = $\frac{5.8}{4}=1.45$

grams of chlorine produced =1.45$\times$71=102.95

As the percent yield is 91 here so finally amount of chlorine gas obtained =$\frac{91}{100}\times102.95=93.7 g$