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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{MnO}_{2}

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{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

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