Question

0.29g of hydrocarbon with vapour density 29 when burnt completely in oxygen produces 448 ml of CO2 at stp. find mass of carbon dioxide formed

Answer 1

vapour density of hydrocarbon is 29
we know,
vapor density = molecular mass/2
molecular mass = 2 × vapor density = 2 × 29 = 58 g/mol

Given,
0.29 g of hydrocarbon produces 448ml of CO2 at STP.

mole of hydrocarbon = given weight/molecular weight
= 0.29/58 = 0.05

mole of CO2 = given volume/22400
= 448/22400 = 0.02

means 0.05 mole of hydrocarbon produces 0.02 mole of CO2 .---(1)

Now see the combustion reaction of hydrocarbon,

CxHy + (x + y/4)O2 ---->xCO2 + (y/2)H2O

here, 1 mole of hydrocarbon produces x mole of CO2,
so, 0.05 mole of hydrocarbon produces 0.05x mole of CO2.---(2)

Compare equation (1) and (2)

0.05x = 0.02
x = 2/5

Let y = 1, then, x/y = 2/5
then, C2H5 is the emperical formula of hydrocarbon .
weight of emperical formula = 2 × 12 + 5 = 29

molecular formula = n × emperical formula

Because, molecular weight = 58g/mol
so, 58 = n × 29
n = 2 , hence, molecular formula will be C4H10

Answer 2

Answer:

refer the answer given above.