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?

7555L

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1586.6L

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850L

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5745.5L

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Solution

The correct option is A
7555L

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?

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?