Question

A sample of KCIO, on decomposition yielded
448 ml of oxygen gas at STP then the weight of
KClO3 originally taken was
(1)0.815g
(2) 1.63g
(3)3.27 (4)2.45g​

Answer 1

Answer:

The weight of  potassium chlorate originally taken was 1.63 g.

Explanation:

Moles of oxygen produced at STP = n

Volume of n moles oxygen gas at STP = 448 mL =0.448 L

1 mole of gas occupies 22.4 L of volume at STP.

Then n moles will ocupy:

$n\times 22.4L=0.448 L$

n = 0.02 mole

$2KClO_3\rightarrow 2KCl+3O_2$

According to reaction 3 moles of oxygen gas are produced from 2 moles of potassium chlorate.

Then 0.02 moles of oxygen will be obtained from:

$\frac{2}{3}\times 0.02 mol=0.01333 mol$

Mass of 0.01333 mole of potassium chlorate:

0.01333 mol × 122.5 g/mol = 1.6329 g ≈ 1.63 g

The weight of  potassium chlorate originally taken was 1.63 g.