Question

What is the extra amount of potassium chlorate to be taken to get $134.4$ L of oxygen at STP by the decomposition of $80\%$ of pure potassium chlorate?

• A. 490 g
• B. 612.5g
• C. 245g
• D. 122.5g

Answer 1

The correct option is B

612.5g

Explanation of correct option:

The reaction for the decomposition of potassium chlorate(${\mathrm{KClO}}_3$) is

$\underset{\mathrm{Postassium}\mathrm{Chlorate}}{2{\mathrm{KClO}}_3}\to \underset{\mathrm{Potassium}\mathrm{Chlorate}}{2\mathrm{KCl}}+\underset{\mathrm{Oxygen}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}$

So, $2\times 122.5$g of ${\mathrm{KClO}}_3$ will produce $3\times 22.4$L of ${\mathrm{O}}_2$.

$<span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>245$g of ${\mathrm{KClO}}_3$= $67.2$L of ${\mathrm{O}}_2$

$<span\; style="font-family:"Cambria\; Math",serif;mso-bidi-font-family:"Cambria\; Math";">\Rightarrow </span>$ $134.4$L of ${\mathrm{O}}_2$= $\frac{245}{67.2}\times 134.4=490$g of ${\mathrm{KClO}}_3$.

Thus,$67.2$ L of ${\mathrm{O}}_2$ should have been produced from $490$g of ${\mathrm{KClO}}_3$.

But the purity of of ${\mathrm{KClO}}_3$ is $80\%$.

If ‘w’ is the weight of the mixture then w = $\frac{490\times 100}{80}=612.5$g.

Thus, $612.5$g of ${\mathrm{KClO}}_3$ is required to produce $134.4$L of ${\mathrm{O}}_2$ at STP.

Hence, the correct answer is option (B).