A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that 16 cm3 of the gas require 48 cm3 of oxygen for complete combustion. This shows that the gas is

Question

A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that 16 cm3 of the gas require 48 cm3 of oxygen for complete combustion. This shows that the gas is (A) only ethane (B) only ethene (C) 1:1 mixtures of two gases (D) some unknown mixture of the two gases

Tardigrade Admin · Accepted Answer

underset16 textc ( textm)3 underset1 textvolCx Hy+ underset16 (x + (y/4)) c m3 underset(x + (y/4)) v o l(x + (y/4)) O2 arrow xCO2+(y/2 H2 O) Now we have 16(x + (y/4))=48 or y = 4 Hence, the hydrocarbon is C2H4 .

Q. A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that 16 $c m^{3}$ of the gas require 48 $c m^{3}$ of oxygen for complete combustion. This shows that the gas is

only ethane

only ethene

1:1 mixtures of two gases

some unknown mixture of the two gases

$16 c (m)^{3} 1 vol C_{x} H_{y} + 16 (x + \frac{y}{4}) c m^{3} (x + \frac{y}{4}) v o l (x + \frac{y}{4}) O_{2} \to x C O_{2} + \frac{y}{2 H _{2} O}$

Now we have $16 (x + \frac{y}{4}) = 48$

or y = 4

Hence, the hydrocarbon is $C_{2} H_{4}$ .

Q. A cylinder of compressed gas that bears no label is supposed to contain either ethane or ethene. Combustion of the sample shows that 16 cm3 of the gas require 48 cm3 of oxygen for complete combustion. This shows that the gas is