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How much potassium chlorate should be heated to produce 2.24L of oxygen at NTP?

seo-qna
Last updated date: 17th Jun 2024
Total views: 372k
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Answer
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Hint: Chemical formula of potassium chlorate is ${\text{KCl}}{{\text{O}}_3}$. First write the equation for the heating of ${\text{KCl}}{{\text{O}}_3}$. Balance the equation. Note that 2 moles of ${\text{KCl}}{{\text{O}}_3}$ produce 3 moles of oxygen.
1 mol = 22.4 L at NTP
3 mol = 67.2 L
Now, 67.2 L oxygen is produced by the heating of 1 mol ${\text{KCl}}{{\text{O}}_3}$
Thus we can find the amount of ${\text{KCl}}{{\text{O}}_3}$ needed to produce 2.24 L oxygen at NTP.

Formula used:
The balanced chemical equation for the heating of ${\text{KCl}}{{\text{O}}_3}$ is:
$2{\text{KCl}}{{\text{O}}_3} \to 2{\text{KCl + 3}}{{\text{O}}_2}$
Also, 1 mol = 22.4 L at NTP

Complete step by step answer:
The balanced chemical equation for the heating of ${\text{KCl}}{{\text{O}}_3}$ is:
$2{\text{KCl}}{{\text{O}}_3} \to 2{\text{KCl + 3}}{{\text{O}}_2}$
Therefore, 2 moles of ${\text{KCl}}{{\text{O}}_3}$ produce 3 moles of oxygen
Now, atomic weights of potassium, chlorine and oxygen are 39, 35.5 and 16 respectively.
molecular weight of 2 moles of ${\text{KCl}}{{\text{O}}_3}$ is 2×(39×35.5×48)=245 gm
Again, 1 mol = 22.4 L at NTP
Therefore, 245 gm of ${\text{KCl}}{{\text{O}}_3}$produces 3×22.4 L of oxygen in NTP
In other words, 3×22.4 L of oxygen produces 245 gm of ${\text{KCl}}{{\text{O}}_3}$ in NTP
2.24 L of oxygen will be produced by
$ = \dfrac{{245 \times 2.24}}{{3 \times 22.4}}$ grams of ${\text{KCl}}{{\text{O}}_3}$
$ = \dfrac{{245}}{{3 \times 10}}$ gm
$ = 8.17$ gm

Hence, 8.17 gm ${\text{KCl}}{{\text{O}}_3}$should be heated in order to produce 2.24 L oxygen in NTP.

Note: Atomic weights of potassium, chlorine and oxygen are 39, 35.5 and 16 respectively.
Also, the balanced chemical equation is $2{\text{KCl}}{{\text{O}}_3} \to 2{\text{KCl + 3}}{{\text{O}}_2}$s when potassium chlorate is heated. ∴ 2 moles of ${\text{KCl}}{{\text{O}}_3}$ produce 3 moles of oxygen.