Question

1gm sample of kclo3 was heated under such conditions that a part of it decomposed according to the equation (1) 2kclo3 2 kcl + 3o2 and remaining underwent change according to the equation. (2) 4kclo3 3 kclo4 + kcl if the amount of o2 evolved was 112 ml at 1 atm and 273 k., calculate the % by weight of kclo4 in the residue.

Answer 1

Answer:

The % by weight of $KClO_4$ in the residue 49.67 %.

Explanation:

Moles of $KClO_3=\frac{1}{122.55 g/mol}=0.0081 mol$  

(1) $2KClO_3\rightarrow 2 KCl + 3O_2$

Moles of oxygen gas at 1 atm and 273 K can be calculated by using ideal gas equation :

$n=\frac{PV}{RT}=\frac{1atm\times 0.112 L}{0.0821 atm L/mol K\times 273 K}$

n = 0.00499 mol = Moles of oxygen gas produced at given conditions

According to reaction(1), 2 moles of $KClO_3$ produces 3 moles of oxygen.

Then 0.00499 mol of oxygen will be produced from:

$\frac{2}{3}\times 0.00499 mol=0.00332 mol$ of $KClO_3$

Moles of $KClO_3$ used in part(1) = 0.00332 mol

(2) $4KClO_3\rightarrow 3KClO_4 + KCl$

Moles of $KClO_3$ used in part(2)=

= 0.0081 mol - 0.00332 mol = 0.00478 mol

According to reaction (2), 4 moles of $KClO_3$ gives 3 moles of $KClO_4$.

Then 0.00478 moles of $KClO_3$ will give:

$\frac{3}{4}\times 0.00478 mol=0.003585 mol$ of$KClO_34$

Mass of 0.003585 moles of$KClO_34$:

0.003585 mole × 138.55 g/mol =0.4967 g

Percentage by weight of $KClO_34$ in the residue:

$\%=\frac{\text{Mass of}KClO_4}{\text{Mass of}KClO_3}$

$\frac{0.4967 g}{1 g}\times 100=49.67\%$

The % by weight of $KClO_4$ in the residue 49.67 %

Answer 2

Answer:

49.67%

no explanation for this question.