What is the weight in g of potassium chlorate | vedclass

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Question

(C) The chemical equations for the reactions are:$(1) \ 2KClO_3 \rightarrow 2KCl + 3O_2$$(2) \ 2CO + O_2 \rightarrow 2CO_2$From equation $(2)$,$2 \ mo

Vedclass · Accepted Answer

$153.12$

Vedclass · Answer

(C) The chemical equations for the reactions are:$(1) \ 2KClO_3 ightarrow 2KCl + 3O_2$$(2) \ 2CO + O_2 ightarrow 2CO_2$From equation $(2)$,$2 \ mol$ of $CO$ requires $1 \ mol$ of $O_2$ ($22.4 \ L$ at $N.T.P$).Therefore,$84 \ L$ of $CO$ requires $\frac{84}{2} = 42 \ L$ of $O_2$.From equation $(1)$,$3 	imes 22.4 \ L$ of $O_2$ is produced by $2 \ mol$ of $KClO_3$.So,$42 \ L$ of $O_2$ is produced by $\frac{2 	imes 42}{3 	imes 22.4} = 1.25 \ mol$ of $KClO_3$.The molar mass of $KClO_3 = 39.1 + 35.5 + 3 	imes 16 = 122.6 \ g/mol$.Weight of $KClO_3 = 1.25 \ mol 	imes 122.6 \ g/mol = 153.25 \ g$.Given the options,the closest value is $153.12 \ g$.

What is the weight in $g$ of potassium chlorate $(KClO_3)$ required to produce the oxygen needed for the complete combustion of $84 \ L$ of $CO$ at $N.T.P$?

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