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Stoichiometry

A sample of g...

Question

A sample of gaseous hydrocarbon occupying $1.12$ litres at NTP when completely burnt in air produced $2.2$ g $C O_{2}$ and $1.8$ g $H_{2} O$ . The amount of volume (in L) of $O_{2}$ at NTP required for its combustion is (as the nearest integer) :

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Solution

Let the formula of hydrocarbon be $C_{a} H_{b}$ .
$C_{a} H_{b} + [a + (b / 4)] O_{2} \to a C O_{2} + (b / 2) H_{2} O$
$∵$ $22.4$ litres of $C_{a} H_{b}$ gives on combustion $=$ $44 a$ g $C O_{2}$
$∴$ $1.12$ litres of $C_{a} H_{b}$ will give on combustion $=$ $\frac{44 a \times 1.12}{22.4} = 2.2 a g$ .
$∴ 2.2 a = 2.2 ∴ a = 1$
Similarly,
$∵$ $22.4$ litres of $C_{a} H_{b}$ gives $=$ $18 \times (b / 2) g H_{2} O$
$∴$ $1.12$ litres of $C_{a} H_{b}$ gives $=$ $18 \times \frac{b}{2} \times \frac{1.12}{22.4} g H_{2} O = \frac{18 \times b}{40} g H_{2} O$
$∴ 1.8 = (\frac{18}{40}) \times b ∴ b = 4$
$∴$ Hydrocarbon is $C H_{4}$
Thus, mass of hydrocarbon taken $=$ $(16 \times 1.12) / 22.4 = 0.8 g$
Also, $O_{2}$ used for combustion of 16 g $C H_{4}$ $=$ $\frac{(a + b)}{4}$ $= 2$ mol
$=$ $2 \times 22.4 l i t r e$
$∴ O_{2}$ used for combustion of $0.8$ g. $C H_{4} = \frac{2 \times 22.4 \times 0.8}{16} = 2.24 l i t r e$
Nearest integer is $2$ L.

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A sample of gaseous hydrocarbon occupying 1.12 litres at NTP when completely burnt in air produced 2.2 g CO2 and 1.8 g H2O. The amount of volume (in L) of O2 at NTP required for its combustion is (as the nearest integer) :

A sample of gaseous hydrocarbon occupying $1.12$ litres at NTP when completely burnt in air produced $2.2$ g $C O_{2}$ and $1.8$ g $H_{2} O$ . The amount of volume (in L) of $O_{2}$ at NTP required for its combustion is (as the nearest integer) :