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Grade: 11
        
A mixture of methane, propane and carbon monoxide contains 36.5% propane by volume. If its 200 ml are burnt in excess of O2, the volume of CO2 formed is :
(a) 173 ml (b) 346 ml (c) 200 ml (d) 519 ml
5 months ago

Answers : (2)

Arun
23520 Points
							
Let the mixture contain a, b and c as C3H8​,
 CH4 (methane) and CO (carbon monoxide) respectively.
a + b + c = 200 mL
a = 36.5% of 200 ml = 36.5100×200 = 73 mL
b + c = 200 mL - 73 mL = 127 mL
Burning in excess of air, following reactions takes place :
C3H8 + 5O2 → 3CO2 + 4 H2O
CH4 + ​ 2O2 → CO2 + 2 H2O
CO + ​ 1/2O2 → CO2 
CO2 formed in terms of mole = 3a +b + c
By putting the value of a we get :
= 3 x 73 mL + (b + c)
= 219 mL + 127 mL
= 346 mL
5 months ago
Vikas TU
10070 Points
							
At constant temperature and pressure:
Volume of C3H8 = 36.5 ml
Volume of CH4 + CO = 200 ml - 36.5 ml = 163.5 ml
Combustion reactions:
C3H8 + 5O2 ==> 3CO2 + 4H2O
CH4 + 2O2 ==> CO2 + 2H2O
CO + 1/2 O2 ==> CO2
On a mole basis, 1 mole C3H8 produces 3 moles CO2 and 1 mole CH4 and CO each form 1 mole CO2
Or, looking at it another way, mls CO2 = 3x36.5 + (200-36.5) = 109.5 + 163.5 = 273 mls of CO2 formed
 
5 months ago
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