Question

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?

Answer 1

ETHANE

Explanation:

Since the gas on burning produces `CO_2 and H_2O`only,then the the burnt gas should be a hydrocarbon.

Let its molecular formula be `C_xH_y` and the balanced equation of the reaction occurred is

`C_xH_y+(x+y/4)O_2->xCO_2+y/2H_2O......(1)`

It is given that at STP 1L of this gas weighs`" " 1.34g.`

So at STP 22.4L of this gas weighs`" "1.34xx22.4g~~30g.`

So molar mass of the gas is `=30g/(mol)`

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 `((4.4xx30)/1.5)`g of CO2and`((2.7.xx30)/1.5)`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 `88/44`mol of CO2and`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 `y/2`mol of H2O

So comparing these two information we can say

`x=2 and y/2=3=>y=6`

Hence the M F of Hydrocarbon is `C_2H_6->` ETHANE