# After burning 1.5g of some gas, 4.4g of CO2 and 2.7g of H2O were produced. What is this gas if at STP 1L of this gas weighs 1.34g?

Aug 26, 2016

ETHANE

#### Explanation:

Since the gas on burning produces $C {O}_{2} \mathmr{and} {H}_{2} O$only,then the the burnt gas should be a hydrocarbon.

Let its molecular formula be ${C}_{x} {H}_{y}$ and the balanced equation of the reaction occurred is

${C}_{x} {H}_{y} + \left(x + \frac{y}{4}\right) {O}_{2} \to x C {O}_{2} + \frac{y}{2} {H}_{2} O \ldots \ldots \left(1\right)$

It is given that at STP 1L of this gas weighs$\text{ } 1.34 g .$

So at STP 22.4L of this gas weighs$\text{ } 1.34 \times 22.4 g \approx 30 g .$

So molar mass of the gas is $= 30 \frac{g}{m o l}$

It is also given that on burning

1.5g of gas produces 4.4g of CO2 and 2.7g of H2O

So burning of 30g of gas

produces $\left(\frac{4.4 \times 30}{1.5}\right)$g of CO2and$\left(\frac{2.7 . \times 30}{1.5}\right)$g of H2O

i.e.burning of 30g of gas produces $88$g of CO2 and $54$g of H2O

i.e.burning of 1 mol of gas produces $\frac{88}{44}$mol of CO2and$\frac{54}{18}$molof H2O

i.e.burning of 1molof gas produces 2 mol of CO2 and 3 mol of H2O

Now from equation (1) we see that
burning of 1 mol of gas produces x mol of CO2 and $\frac{y}{2}$mol of H2O

So comparing these two information we can say

$x = 2 \mathmr{and} \frac{y}{2} = 3 \implies y = 6$

Hence the M F of Hydrocarbon is ${C}_{2} {H}_{6} \to$ ETHANE