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`        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

Arun
23520 Points
```							Let the mixture contain a, b and c as C3H8​, CH4 (methane) and CO (carbon monoxide) respectively.a + b + c = 200 mLa = 36.5% of 200 ml = 36.5100×200 = 73 mLb + c = 200 mL - 73 mL = 127 mLBurning in excess of air, following reactions takes place :C3H8 + 5O2 → 3CO2 + 4 H2OCH4 + ​ 2O2 → CO2 + 2 H2OCO + ​ 1/2O2 → CO2 CO2 formed in terms of mole = 3a +b + cBy 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 mlVolume of CH4 + CO = 200 ml - 36.5 ml = 163.5 mlCombustion reactions:C3H8 + 5O2 ==> 3CO2 + 4H2OCH4 + 2O2 ==> CO2 + 2H2OCO + 1/2 O2 ==> CO2On a mole basis, 1 mole C3H8 produces 3 moles CO2 and 1 mole CH4 and CO each form 1 mole CO2Or, 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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