Higher Chemistry Volume of Gas?
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Higher Chemistry Volume of Gas?

[From: ] [author: ] [Date: 14-05-07] [Hit: ]
V=nRT/p =0.01604mol * 0.0821atm L/mol K * 298.15K / 1atm =0.......
Since 2 moles ethylene yields 4 moles of CO2 (that's the gas the question is referring to), the mole ratio is 1:2 for ethylene : CO2.

Using volumes instead of moles, 100 cm3 ethylene will yield 200 cm3 of CO2. The same 1:2 ratio applies.

Your confusion comes from looking at O2 instead of CO2. The question asks how much gas is PRODUCED, not how much gas is USED.

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2C2H2(g) + 5O2(g) ----> 4CO2(g) + 2H2O(l)
V(C2H2)=100cm3
we need the number of moles of ethyne
n=PV/RT= ( 1atm * 0.1L ) / ( 0.0821 atm L/mol K * 298.15K )=4.08*10^-3

but we have 4 of CO2 so we have 4 * 4*10^-3= 0.1604 moles of CO2
now we need the volume of CO2
V=nRT/p =0.01604mol * 0.0821atm L/mol K * 298.15K / 1atm =0.393L= 393 cm3
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