Question

How much quantity of zinc in grams will have to be reacted with excess of dilute HCl solution to produce sufficient hydrogen gas for completely reacting with the oxygen obtained by decomposing $5.104$ g of potassium chlorate?

• A. $12$
• B. $45$
• C. $5$
• D. $7.85$

Answer 1

The correct option is D $7.85$

$2KCl{O}_3\left(s\right)\to 3O_2\left(g\right)+2KCl$

As $2$ mol of $KCl{O}_3$ gives $3$ mol of $O_2$. It means $245.2$ g of $KCl{O}_3$ gives $3$ mol of $O_2$.

So, $5.104$ g ( $0.04$ moles ) of $KCl{O}_3$ will give , $\frac{3}{2}\times 0.04=0.06$ moles of $O_2$.

Now , $Zn+2HCl\to ZnC{l}_2+H_2$.

$65.4$ g of $Zn$ gives $1$ mol of $H_2$.

$2H_2+O_2\to 2H_{2}O$

$1$ mol of $O_2$ reacts with $2$ mol of $H_2$.

So $0.06$ mol of $O_2$, produced by $KCl{O}_3$ decomposition, will react with $0.12$ mol of $H_2$.

$1$ mol $H_2$ produced by $65.4$ g of $Zn$, so $0.12$ mol $H_2$ is produced by $65.4\times 0.12=7.85$ g of $Zn$ is required.