Answer to Question #116328 in General Chemistry for Nolan Ryang

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 =$\frac{91}{100}\times102.95=93.7 g$