Question

calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at stp, molar mass of potassium chlorate is 122.5 g/ mol.

Answer 1

Answer: 24.5 grams of potassium chlorate.

Explanation:

According to avogadro's law, 1 mole of every substance occupy 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

According to stochiometry:

2 moles of $KClO_3$ produce 3 moles of $O_2$  

Thus $3\times 22.4L=67.2L$ of $O_2$ is produced from $2\times 122.5=245g$ of $KClO_3$

6.72 L of $O_2$ is produced from =$frac{245}{67.2}\times 6.72=24.5g$of $KClO_3$

Thus mass of potassium chlorate required to liberate $6.72dm^3$ of oxygen at STP will be 24.5 grams.

Answer 2

Explanation:

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