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Limiting Reagent or Reactant

MnO2 +4HCl⟶ M...

Question

$M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}$
$2$ moles $M n O_{2}$ reacts with $4$ moles of $H C l$ to form $11.2$ L $C l_{2}$ at STP. Thus, percent yield of $C l_{2}$ is:

A

25 %

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B

50 %

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C

100 %

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D

75 %

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Solution

The correct option is B $50 %$
The reaction is as follows:
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}$
$2$ mol $4$ mol (limiting reagent)
So, according to equation, $22.4$ L of $C l_{2}$ should be formed but, only $11.2$ L forms.
Therefore, $%$ yield is $11.2 \times \frac{100}{22.4} = 50 %$ .

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Similar questions

Q. Select correct alternate. Only one alternate is correct.
In the following reaction :

M n O_{2} + 4 H C l ⟶ M n C l_{2}, + 2 H_{2} O + C l_{2}

2

M n O_{2}

reacts witb 4 moles of HCl to form

11.2 L C l_{2}

, at STP. Thus, percent yield of

C l_{2}

0.80

M n O_{2}

146

H C l

react, then number of moles of

C l_{2}

formed is (as nearest integer):

M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O

0.80

M n O_{2}

146

H C l

react in the following manner:

M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O

M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O

0.86

M n O_{2}

48.2

H C l

react, how many grams of

C l_{2}

will be produced? Report your answer to the nearest rounded integer.

0.80

M n O_{2}

146

H C l

M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O

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