calculate the mass of potassium chlorate required to liberate 6.72 dm�

1.	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» 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 $KClO_3$ $6.72dm^3$ Thus mass of potassium chlorate required to liberate of oxygen at STP will be 24.5 grams.

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»

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 $KClO_3$

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