Calculate the heat change during the reaction $24$ g C and $128$ g S following :the change $C + S_{2} \to C S_{2}$ ; $Δ$ H $= 22$ K cal.

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Solution

$C + S_{2} ⟶ C S_{2} Δ H = 22 k c a l$
$1$ mole of $C = 12 g$
$1$ mole of $S_{2} = 64 g$
Therefore,
$24 g$ of $C = 2 moles$
$128 g$ of $S_{2} = 2 moles$
Given that heat change during the reaction of $12 g$ of $C$ and $64 g$ of $S$ is $22 k c a l$
Therefore
$2 \times [C + S_{2} ⟶ C S_{2} Δ H = 22 k c a l]$
$2 C + 2 S_{2} ⟶ 2 C S_{2} Δ H = 44 k c a l$
Hence the heat change during the reaction $24 g$ of $C$ and $128 g$ of $S$ is $44 k c a l$ .

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Calculate the heat change during the reaction 24g C and 128g S following :the change C+S2→CS2; ΔH=22K cal.

Calculate the heat change during the reaction $24$ g C and $128$ g S following :the change $C + S_{2} \to C S_{2}$ ; $Δ$ H $= 22$ K cal.