Question

a sample of KClO3 on decomposition fielded for 448 ml of Oxygen gas at STP then the weight of KClO3 original you taken was

Answer 1

Answer:-

Oxygen = 448 ml

moles produced =O2

22400 =448

=0.02moles

→2KCl+3O 2

POAC in Oxygen Atoms

Oxygen in KClO3 oxygen in O2

KClO3) = 3 moles of O2

(=3×0.022× 122.5x)

=0.06

x=0.03×122.5

x=3.675g

Answer=3.675g

Answer 2

Answer: 1.6 grams.

Explanation:

To calculate the moles, we use the equation:

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

For $O_2$

Given volume = 448 ml

Molar volume = 22.4 L= 22400 ml

Putting values in above equation, we get:

$\text{Moles of}O_2 =\frac{448}{22400}=0.02moles$

$2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

3 moles of $O_2$ are produced from 2 moles of $KClO_3$

0.02 moles of $O_2$ are produced from =$\frac{2}{3}\times 0.02=0.013$ moles of $KClO_3$

mass of $KClO_3=moles\times {\text {molar mass}}=0.013\times 122.5=1.6g$

The weight of $KClO_3$  original you taken was 1.6 grams.