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.
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
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