Question

Calculate the amount of `KClO_(3)` needed to supply sufficient oxygen for burning 112 L of CO gas at N.T.P.

Answer 1

The chemical equations for the above reactions are :
`{:(" "2KClO_(3)overset("Heat")(rarr)2KCl+3O_(2)),(" "2CO+O_(2)rarr2CO_(2)"]"xx3),(bar(undersetunderset(=245g)(2xx(39+35.5+48))(2KClO_(3))+undersetunderset(134.4L)(6xx22.4)(6CO)rarr3KCl+6CO_(2))):}`
134.4 L of CO require `KClO_(3)=245` g
1.0 L of CO require `KClO_(3) = (245)/(134.4)g`
112 L of CO require `KClO_(3)=(245)/(134.4)xx112 = 204.167` g.