For the following reaction given bellow, calculate how much heat will be released when 8 g of hydrogen reacts. (1 mole of $H_{2}$ weighs approximately 2 grams.)
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$
$Δ H = - 115.60 k c a l$

57.8 kcal

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115.6 kcal

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173.4 kcal

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231.2 kcal

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462.4 kcal

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Solution

The correct option is D 231.2 kcal
$2 H_{2} + O_{2} \to 2 H_{2} O$ ; $Δ H_{r e a c t i o n} = - 115.6 k c a l$
In this reaction the weight of $H_{2}$ gas = 4 g.
Now, for 8 g of $H_{2}$ gas the above reaction and enthalpy of the reaction must be multiplied by 2.
$2 \times [2 H_{2} + O_{2} \to 2 H_{2} O];$ $2 \times Δ H_{r e a c t i o n} = 2 \times (- 115.6) = 231.2 k c a l$

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