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Stoichiometric Calculations

Given the rea...

Question

Given the reaction at STP: $M g (s) + 2 H C l (a q) \to M g {C l}_{2} (a q) + H_{2} (g)$ , how many liters of $H_{2} (g)$ can be produced from the reaction of 12.15 g Mg and excess ?

2.0 L

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4.0 L

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11.2 L

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22.4 L

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44.8 L

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Solution

The correct option is A 11.2 L
At STP :
$M g + 2 H C l \to M g C l_{2} + H_{2}$
As per reaction, 1 mol ( 24.30 g) of $M g$ produces 1 mol ( 22.4 L) gas.
So, $\frac{1}{2} m o l$ or $12.15 g$ of Mg will produce $\frac{1}{2} m o l$ of $H_{2}$ gas which has volume $(11.2 L)$ at STP.

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