Question

Kclo3 decomposes to kcl and o2 if the voume of o2 obtained in this reaction is 1.12 lit at STP weight of kcl formed in this reaction is

Answer 1

The decomposition of KclO3 is follows:

$KClO_3    2 KCl   +  3 O_2$

2 mole of KCl is obtained form 2 mol KClO3Volume of O2 = 1120 mL

As, PV = nRT

$n_O_2 =  PV /RT$

$= (1) * 1.12 / 0.0831 * ( 273.15 K)$

= 0.049 mol

Mole of KCl = (2 /3) (0.049) = 0.032 mol

$m /M = n = 0.032$

m/ 74 = 0.032

m = 2.43 g

Thus, 2.43 g potassium chloride is produced.

Answer 2

Answer: Amount of KCl formed at STP will be 2.4585 g

Explanation:

At STP:

According to mole concept:

22.4 L of volume is occupied by 1 mole of gas.

So, 1.12 L of volume will be occupied by = $\frac{1}{22.4}\times 1.12=0.05mol$ of oxygen gas.

The chemical equation for the decomposition of $KClO_3$ is given by:

$2KClO_3\rightarrow 2KCl+3O_2$

By Stoichiometry of the reaction:

When 3 moles of oxygen gas is formed, then 2 moles of potassium chloride is also forming.

So, when 0.05 moles of oxygen gas will be formed, then $=\frac{2}{3}\times 0.05=0.033$ moles of potassium chloride will form.

To calculate the amount of potassium chloride formed, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of potassium chloride = 0.033 mol

Molar mass of potassium chloride = 74.5 g/mol

Putting values in above equation, we get:

$0.033=\frac{\text{Mass of KCl}}{74.5g/mol}\\\\\text{Mass of KCl}=2.4585g$

Hence, amount of KCl formed at STP will be 2.4585 g