The heat evolved in the combustion of methane is given by the following equations $CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l);\ \mathrm{\Delta}H = - 890.3\ kJ$ How many grams of methane would be required to produce 44.15 kJ of heat of combustion?

Question

Dr. Abubacker · Accepted Answer

(b)
$CH_{4}$ required $= \frac{445.15 	imes 16}{890.3} = 8$ g

Q. The heat evolved in the combustion of methane is given by the following equations $CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l);\ \mathrm{\Delta}H = - 890.3\ kJ$ How many grams of methane would be required to produce 44.15 kJ of heat of combustion?

Explanation

Reviewed by Dr. Abubacker

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