Calculate the amount of $C O_{2}$ produced (in grams) when 1 mol of $C H_{4}$ is burnt:

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Solution

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
When $1$ mole of $C H_{4}$ is burnt, $1$ mole of $C O_{2}$ is produced and 1 mole $C O_{2} = 12 + 2 \times 16 = 44 g$ .

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The percentage by volume of $C_{3} H_{8}$ in a mixture of $C_{3} H_{8}$ , $C H_{4}$ and CO is 36.5. Calculate the volume of $C O_{2}$ produced when 100 mL of mixture is burnt in excess of $O_{2}$ .

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Calculate the amount of CO2 produced (in grams) when 1 mol of CH4 is burnt:

CH4+2O2→CO2+2H2OWhen 1 mole of CH4 is burnt, 1 mole of CO2 is produced and 1 mole CO2=12+2×16=44g.

Calculate the amount of $C O_{2}$ produced (in grams) when 1 mol of $C H_{4}$ is burnt:

$C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
When $1$ mole of $C H_{4}$ is burnt, $1$ mole of $C O_{2}$ is produced and 1 mole $C O_{2} = 12 + 2 \times 16 = 44 g$ .