Question

1 .
A sample of KCIO,3 on decomposition yielded
448 mL of oxygen gas at STP, then the weight of
KCIO3, originally taken was​

Answer 1

Volume of Oxygen produced = 448 ml

We know that :-

Number of moles of O2 produced will be as follows :-

$= \frac{448}{22400} \\ \\ \\ = 0.02moles$

$\rule{200}{2}$

Balanced Chemical Equation has :-

$\boxed{\bold{\sf{\red{2KClO_3 \rightarrow 2KCl + 3O_2}}}}$

$\rule{200}{2}$

Applying POAC on Oxygen atoms in the equation :-

Moles of Oxygen in KClO3 is same as number of moles of oxygen in O2

Thus,

2 (moles of KClO3) = 3(moles of O2)

Hence,

$2 \times \frac{weight \: of \: kclo3}{molecular \: weight \: of \: kclo3} = 3 \times 0.02 \\ \\ \\2 \times \frac{x}{122.5} = 0.06 \\ \\ \\x = 0.03 \times 122.5 \\ \\ \\x = 3.675g$

$\rule{200}{2}$

Hence,

Originally, 3.675 grams of potassium chlorate was taken.

Answer 2

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