Question

15 ml of a gaseous hydrocarbon was required for complete combustion in 357ml of air (21% of oxygen by volume) and the gaseous products occupied 327 ml (all volumes being measured at NTP). What is the formula of the hydrocarbon?

• A. C3H8
• B. C4H8
• C. C5H10
• D. C4H10

Answer 1

Answer: (A)

C3H8

Answer 2

(1)

$C_{x}H_{y}\text{ + }\text{O}_{2}\ \overset{\quad\quad}{\rightarrow}\text{ }\text{x}_{\text{C}\text{O}_{2}}\text{+ }\frac{y}{2}\text{ + }\text{H}_{2}\text{O }15ml\frac{357 \times 21}{100}ml$

75 ml

$\left( x + \frac{y}{4} \right) \times 15 = 75$

$x + \frac{y}{4} = \frac{75}{15}$

$x + \frac{y}{4} = 5$

$x + \frac{y}{4} = 5$

$3 + \frac{y}{4} = 5$

15 x + 15x + 282 = 327

y = 8

x = 3

Formula = C3H8