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Question

# The percentage volume of C3H8 in a mixture of C3H8, CH4 and CO is 36.5. The volume of CO2 produced when 100 mL of mixture is burnt in excess of O2 is:

A
273 mL
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B
183 mL
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C
173 mL
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D
193 mL
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Solution

## The correct option is B 273 mLLet the volume of mixture of all gases be 100 unit.The reaction of combustion as follows:C3H8+CH4+CO+15/2O2→5CO2+6H2OIt means 5 mol of CO2 is produced when 1 mol C3H8 is burnt in a given mixture, along with 1 mol each of CH4 and CO.As 22400 mL of C3H8 contains 1 mol, then 36.5 mL contains =122400×36.5 moles1 mol of C3H8 produces 3 mol of CO2 (2 moles from CH4 and CO)Then, 122400×36.5 mol produces 36.522400×3 moles of CO2.Now CH4 + CO combinely must have (122400×100)−(122400×36.5) moles, in 100 mL mixtureNow as per the above molar equation for all gases of the mixture, mixture gives 5 mol of CO2.so 1 mol of each CH4 and CO must give 2 mol of CO2.so from 100 mL mixture of CH4+CO will give =10022400×2−122400×36.5 molesNow total mol from the mixture, CO2 moles produced=10022400×2−122400×36.5+36.522400×3=27322400 =0.0121875molesAs 1 mol CO2 has 22400 mL volume.So 0.0121875 mol have volume=0.0121875×22400=273mLHence, option A is correct.

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