You visited us 1 times! Enjoying our articles? Unlock Full Access!

Molarity

MnO2 + 4HCl⟶ ...

Question

$M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O$
If $0.86$ 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.

Open in App

Solution

The balanced chemical equation is as given below:
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + 2 H_{2} O + C l_{2}$
$1 m o l 4 m o l$
$0.86 m o l 48.2 g = 1.32 m o l$
$H C l$ required by $0.86$ mol $M n O_{2} = 3.44 m o l$
but $H C l$ taken $= 1.32 m o l$
Thus, $H C l$ is the limiting reactant.
Thus, $C l_{2} f o r m e d = \frac{1}{4} \times 1.32$
$= 0.33 m o l C l_{2}$
$= 23.43 g$
Hence, $23$ g of chlorine will be produced.

Suggest Corrections

Similar questions

0.80

M n O_{2}

146

H C l

C l_{2}

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

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

(M n O_{2})

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

M n O_{2}

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

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}

C l_{2}

Join BYJU'S Learning Program