$2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$

Calculate the mass of $K C l O_{3}$ required to produced 6.72 litre of $O_{2}$ at S.T.P. [K=39, Cl=35.5, O=16].

24.5 gm

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30 gm

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50 gm

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60 gm

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Solution

The correct option is D 24.5 gm
$2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$
Moles of oxygen $= \frac{6.72}{22.4} = 0.3$

2 moles of $K C l O_{3}$ will produce 3 moles of $O_{2}$ .

0.3 moles of oxygen will be produced by $\frac{2}{3} \times 0.3 = 0.2$ moles of chlorate.

Hence, mass of $K C l O_{3}$ required $= 0.2 \times 122.5 = 24.5$ g .

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Similar questions

2 K C l O_{3} △ - \to 2 K C l + 3 O_{2}

4.9 g

[K = 39; O = 16; C l = 35.5]

2 K C l O_{3} Δ \to 2 K C l + 3 O_{2}; C + O_{2} Δ \to C O_{2}

K C l O_{3}

O_{2}

{K C l O}_{3}

{K C l O}_{3}

K C l

{2 K C l O}_{3} Δ - \to {2 K C l + 3 O}_{2}

[K = 39, C = 25.5

O = 16]

K C l O_{3}

(K = 39, C l = 35.5, O = 16)

a) What is percentage composition of potassium chlorate (KClO₃).

Given:Atomic mass Of K=39,Cl=35.5andO=16

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