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?

Question

Noah Aldrich · Accepted Answer

ETHANE

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 known that one liter of this gas weighs 1.34 grams at STP. So at STP 22.4L of this gas weighs#" "1.34xx22.4g~~30g.# So molar mass of the gas is #=30g/(mol)# Also, it is stated that upon burning 4.4g of CO2 and 2.7g of H2O are produced from 1.5g of gas. Hence, burning 30 grams of gas produces #((4.4xx30)/1.5)#g of CO2and#((2.7.xx30)/1.5)#g of H2O For example, burning 30 grams of gas yields 88 grams of CO2 and 54 grams of H2O. i.e.burning of 1 mol of gas produces #88/44#mol of CO2and#54/18#molof H2O For example, burning one mol of gas yields two mol of CO2 and three 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 Thus, by contrasting these two pieces of data, we can say #x=2 and y/2=3=>y=6# Hence the M F of Hydrocarbon is #C_2H_6-># ETHANE

Henry Antrim · Answer

undefined The gas is methane (CH4).