Question

Calculate the mass of kclo3 thar will liberate 11.2 l of o2 at stp

Answer 1

Answer: 40.83 g

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and weighs equal to the molecular mass.

$2KClO_3\rightarrow 2KCl+3O_2$  

According to stochiometry:

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

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

67.2 L of $CO_2$ is produced by 245 g of $KClO_3$

11.2 L of $CO_2$ is produced by =$\frac{245}{67.2}\times 11.2=40.83g$

Answer 2

At STP , Volume of 1 mol = 22.4 L 2KClO3 → 2KCl + 3O2

Molar mass of KClO3 = 39 + 35.5 + 3××16 = 122.5 g

22.4 L of O2 produced at NTP = 122.5 g of KClO3

11.2 L of O2 produced = 122.5/22.4×11.2/122.524.8×11.2 = 40.83 g of KClO3 PLEASE BRAINLIEST