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Limiting Reagent or Reactant

If 0.80 mol...

Question

If $0.80$ mol of $M n O_{2}$ and $146$ g of $H C l$ react, then :
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O$

0.80

mol of

C l_{2}

is formed

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0.80

mol of

H C l

remains unreacted

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M n O_{2}

is completely reacts

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M n O_{2}

is the limiting reactant

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Solution

The correct options are
A $0.80$ mol of $C l_{2}$ is formed
B $0.80$ mol of $H C l$ remains unreacted
C $M n O_{2}$ is completely reacts
D $M n O_{2}$ is the limiting reactant
The balanced reaction is given below:
$M n O_{2} 1 m o l + 4 H C l 4 m o l ⟶ M n C l_{2} + C l_{2} 1 m o l + 2 H_{2} O$
Initial $0.8$ mol $4$ mol $0$ mol
After $0$ mol $4 - 3.2 = 0.8$ mol $0.8$ mol
$146 g H C l = \frac{146}{36.5} = 4 m o l$
Thus, $M n O_{2}$ (less) is limiting reagent.
$H C l$ (more) is in excess.
Amount of product is decided by $M n O_{2}$ .
Therefore, $C l_{2} f o r m e d = 0.8 m o l$ and $H C l u n r e a c t e d = 0.8 m o l$ .

Suggest Corrections

Similar questions

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. 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 :

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. 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

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

0.86

mole of

M n O_{2}

and

48.2

g of

H C l

react, how many grams of

C l_{2}

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

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If 0.80 mol of MnO2 and 146 g of HCl react, then :MnO2+4HCl⟶MnCl2+Cl2+2H2O

If $0.80$ mol of $M n O_{2}$ and $146$ g of $H C l$ react, then :
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O$