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Limiting Reagent or Reactant

If 0.80 mol...

Question

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

moles of

C l_{2}

is formed

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0.80

moles of

H C l

remains unreacted

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M n O_{2}

is completely reacted

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M n O_{2}

is the limiting reactant

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Solution

The correct options are
A $0.80$ moles of $C l_{2}$ is formed
B $0.80$ moles of $H C l$ remains unreacted
C $M n O_{2}$ is completely reacted
D $M n O_{2}$ is the limiting reactant
$146 g H C l = \frac{146}{36.5} = 4 m o l$

The balanced reaction is as follows:

$M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O$
$0.8$ $4$ $0$ moles (Initial)
$0$ $0.8$ $0.8$ moles (Final)

Thus, $M n O_{2}$ (less) is a limiting reagent and $H C l$ is in excess.

Amount of product is decided by $M n O_{2}$ .

$C l_{2}$ $= 0.8 m o l$ .

$H C l$ unreacted $= 4 - 3.2 = 0.8 m o l$ .

Hence, option A, B, C and D are correct.

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