Question

Compute the mass in grams of KClO3 necessary to produce 67.2 liters of oxygen at S. T. P. according to the reaction
2KClO3 (s) —> 2KCl(s) +3O2 (g)
[Molecular weight of KClO3= 122.5]

Answer 1

Answer: 245 grams

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 Liters 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 stoichiometry, 3 moles of $O_2$ are produced from 2 moles of $KClO_3$

Thus $3\times 22.4=67.2L$ of  $O_2$ are produced from 2 moles of $KClO_3$

Mass of $KClO_3=moles\times {\text {Molar mass}}=2\times 122.5=245g$

Thus 245 grams of $KClO_3$ necessary to produce 67.2 liters of oxygen at S. T. P.

Answer 2

Answer:

245g

Explanation:

1mole= 22.4L at STP

2mole of kclo3 gives 3 mole of o2

3× 22.4L = 67.2 L

mass = no. of moles × molar mass

mass= 2×122.5

mass= 245g