How much heat in $KJ$ is released during the combustion of $112 \ L$ of water gas at $S.T.P.$? (Water gas is an equimolar mixture of $H_2$ and $CO$)
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}; \Delta H = -241.8 \ KJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -283 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}; \Delta H = -241.8 \ KJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -283 \ KJ$
- A$241.8$
- B$283$
- C$1312$
- D$1586$
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