What volume of air containing $21$ % of oxygen by volume is required to completely burn one kg of carbon that contains $15$ % of non-combustible substance?

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Solution

Combustible part out of $1 k g$ = $850 g m$

$850 g$ of carbon= $70.83 m o l e s$
1 mole of C requieres one mole of oxygen for complete combustion:
$C + O_{2} \to C O_{2}$

Vol. of oxygen needed for $70.83 m o l e$ of carbon = $70.83 \times 22.4 = 1586.6 L$

$∴$ Vol. of air that can supply $1586.6 L$ of oxygen = $\frac{1586.6 \times 100}{21}$ = $7555.5 L$

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What volume of air containing 21% of oxygen by volume is required to completely burn one kg of carbon that contains 15% of non-combustible substance?

Combustible part out of 1 kg = 850gm 850g of carbon= 70.83moles 1 mole of C requieres one mole of oxygen for complete combustion: C+O2→CO2 Vol. of oxygen needed for 70.83 mole of carbon = 70.83 × 22.4 = 1586.6L ∴ Vol. of air that can supply 1586.6 L of oxygen = 1586.6 × 10021 = 7555.5L

What volume of air containing $21$ % of oxygen by volume is required to completely burn one kg of carbon that contains $15$ % of non-combustible substance?