Question

If potassium chlorate is 80 pure then 48g of oxygen would be produced from

Answer 1

2KClO3=2KCl+3O2.So 245g of KClO3 produces 96g of oxygen.So 122.5g of KClO3 produces 48g of oxygen.This is when it’s 100% pure.But it’s 80% pure.So therefore 152.12 g of KClO3 is required

Answer 2

Answer: 153.125 grams

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

$2KClO_3\rightarrow 2KCl+3O_2$

According to stochiometry:

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

Thus $3\times 32=96g$ of $O_2$ is produced from $2\times 122.5=245g$ of $KClO_3$

48 g of $O_2$ is produced from =$frac{245}{96}\times 48g=122.5g$of $KClO_3$

But as % purity of $KClO_3$ is 80%, the amount of $KClO_3$ required will be is=$\frac{100}{80}\times 122.5=153.125g$

Thus 48g of oxygen would be produced from 153.125 g of $KClO_3$