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Stoichiometric Calculations

5.0 g of KC...

Question

$5.0$ g of $K C l O_{3}$ gave $0.03$ mol of $O_{2}$ . Hence, percentage purity of $K C l O_{3}$ is :
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$

A

49 %

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B

50 %

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C

95 %

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D

98 %

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Solution

The correct option is B $50 %$
$2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}$
$5$ g
$\frac{5}{122.5} = 0.04$ mol
So, according to equation, if reactant are $100 %$ pure, then $0.06$ mol of $O_{2}$ should be formed, but only $0.03$ mol formed.
So, $%$ purity is $0.03 \times \frac{100}{0.06} = 50 %$ .

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Similar questions

A l

K C l O_{3}

react together to form

A l_{2} O_{3}

according to the following reactions :

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A I_{2} O_{3}

4

K C l O_{3}

50 %

pure) on reaction with excess of Al forms _______ moles of

A l_{2} O_{3}

A l

K C l O_{3}

react together to form

A I_{2} O_{3}

according to the following reaction:

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}

4

K C l O_{3}

50 %

pure) on reaction with excess of

A l

A l_{2} O_{3}

A l

K C l O_{3}

react together to form

A l_{2} O_{3}

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A l_{2} O_{3}

4

K C l O_{3}

50 %

pure) on reaction with excess of

A l

forms x moles of

A l_{2} O_{3}

. The value of x (as nearest integer) is:

K C l O_{3}

react together to form

A I_{2} O_{3}

2 K C l O_{3} ⟶ 2 K C l + 3 O_{2}

4 A l + 3 O_{2} ⟶ 2 A I_{2} O_{3}

4 m o l e s o f K C l O_{3}

(50% pure) on reaction with excess of

A l

x

A I_{2} O_{3}

.

M n O_{2} + 2 K C l O_{3} ⟶ K_{2} M n O_{4} + C l_{2} + 2 O_{2}

C l_{2} + K M n O_{4} ⟶ 2 K C l + M n O_{2} + O_{2}

Each reaction takes place to the extent of

50 %

11.2

O_{2}

at STP is obtained from

K C l O_{3}

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