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Mass, Volume and Mole Relationship

How many gram...

Question

How many grams of ${MnO}_{2}$ are needed to get 36.2g of ${MnCl}_{2}$ from the equation given below?
${MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g$
25

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Solution

The correct option is A 25
In the given equation 87g of ${MnO}_{2}$ gives 126g of ${MnCl}_{2}$ .

To produce 1g of ${MnCl}_{2}$ the amount of ${MnO}_{2}$ required = $\frac{87}{126}$ g

Hence, to produce 36.2 g of ${MnCl}_{2}$ the amount of ${MnO}_{2}$ required = $\frac{87}{126} \times 36.2$ = 25 (approx.)

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Similar questions

{MnO}_{2}

{MnCl}_{2}

{MnO}_{2} (s) + 4 HCl (aq) \to {MnCl}_{2} (aq) + 2 H_{2} O (l) + {Cl}_{2} (g) 87 g 36.5 g 126 g 18 g 71 g

69.6

M n O_{2}

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

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

C l_{2}

(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}

1

H C l

1

M n O_{2}

C l_{2}

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

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