The heat evolved in the combustion of methane | vedclass

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(B) From the given equation,$890.3 \, kJ$ of heat is liberated by the combustion of $1 \, mol$ of $CH_4$.Therefore,$1 \, kJ$ of heat is liberated by $

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$8$

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(B) From the given equation,$890.3 \, kJ$ of heat is liberated by the combustion of $1 \, mol$ of $CH_4$.Therefore,$1 \, kJ$ of heat is liberated by $\frac{1}{890.3} \, mol$ of $CH_4$.For $445.15 \, kJ$ of heat,the amount of $CH_4$ required is $\frac{445.15}{890.3} \, mol = 0.5 \, mol$.The molar mass of $CH_4$ is $16 \, g/mol$.Mass of $CH_4 = 0.5 \, mol 	imes 16 \, g/mol = 8 \, g$.

The heat evolved in the combustion of methane is given by the following equation:
$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)};$
$\Delta H = -890.3 \, kJ$
How many grams of methane would be required to produce $445.15 \, kJ$ of heat of combustion?

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The heat evolved in the combustion of methane is given by the following equation:$CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)};$$\Delta H = -890.3 \, kJ$How many grams of methane would be required to produce $445.15 \, kJ$ of heat of combustion?