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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 number of moles of $C l_{2}$ formed is (as nearest integer):
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O$

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Solution

$146$ g $H C l = \frac{146}{36.5} = 4$ mol

$M n O_{2} 1 m o l + 4 H C l 4 m o l ⟶ M n C l_{2} + C l_{2} 1 m o l + 2 H_{2} O$
Initial $0.8$ mol $4$ mol $0$ mol
After $0$ mol $0.8$ mol $0.8$ mol

Thus, $M n O_{2}$ (less) is the limiting reagent and $H C l$ is in excess.
The amount of product is decided by $M n O_{2}$ .
$C l_{2}$ formed $= 0.8$ mol $\approx 1$ mol.

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Similar questions

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

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.

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}

4

H C l

11.2

C l_{2}

at STP. Thus, percent yield of

C l_{2}

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

1

H C l

1

M n O_{2}

is heated together, the maximum weight of

C l_{2}

gas evolved will be:

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