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?

1 Answer
Aug 26, 2016

Answer:

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