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Molarity

Select correc...

Question

Select correct alternate. Only one alternate is correct.
In the following reaction :
$M n O_{2} + 4 H C l ⟶ M n C l_{2}, + 2 H_{2} O + C l_{2}$
$2$ moles $M n O_{2}$ reacts witb 4 moles of HCl to form $11.2 L C l_{2}$ , at STP. Thus, percent yield of $C l_{2}$ is :

25%

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50%

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100%

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75%

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Solution

The correct option is B 50%
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}$
2 mole 4 mole (limiting reagent)
So, according to equation 22.4 lit of $C l_{2}$ should be form but as 11.2 lit forms only so % yield is $11.2 * (100 / 22.4) = 50$ %.

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Similar questions

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

C l_{2}

at STP. Thus, percent yield of

C l_{2}

is:

Q. If

0.80

moles of

M n O_{2}

and

146

g of

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

, then :

0.80

mol of

M n O_{2}

and

146

g of

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

Q. Assertion :One equivalent of

M n O_{2}

reacts with 2 equivalents of

H C l

in the reaction,

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

. Reason: One equivalent of

M n O_{2}

reacts with one equivalent of

H C l

Q. If

0.80

mol of

M n O_{2}

and

146

g of

H C l

react, then :

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

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Select correct alternate. Only one alternate is correct.In the following reaction : MnO2+4HCl⟶MnCl2,+2H2O+Cl22 moles MnO2 reacts witb 4 moles of HCl to form 11.2LCl2, at STP. Thus, percent yield of Cl2 is :

The correct option is B 50%MnO2+4HCl⟶MnCl2+2H2O+Cl22 mole 4 mole (limiting reagent)So, according to equation 22.4 lit of Cl2 should be form but as 11.2 lit forms only so % yield is 11.2∗(100/22.4)=50 %.

Select correct alternate. Only one alternate is correct.
In the following reaction :
$M n O_{2} + 4 H C l ⟶ M n C l_{2}, + 2 H_{2} O + C l_{2}$
$2$ moles $M n O_{2}$ reacts witb 4 moles of HCl to form $11.2 L C l_{2}$ , at STP. Thus, percent yield of $C l_{2}$ is :

The correct option is B 50%
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}$
2 mole 4 mole (limiting reagent)
So, according to equation 22.4 lit of $C l_{2}$ should be form but as 11.2 lit forms only so % yield is $11.2 * (100 / 22.4) = 50$ %.