solve this by using gas equation law :
Q. what volume of oxygen at stp is required to affect the combustion of 11 litres of ethylene at 273degree celcious and 380 mm hg pressure ?
answer by verifying because i am not able to answer this .
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The balanced chemical equation for the proposed reaction is:
C2H4 + 3O2 → 2H2O + 2CO2
3 mols of O2 are required to fully combust 1 mol of C2H4
11 litres required 3/1 * 11=33 litre's of oxygen Now , p1= 760mm, T1=273k , v1=? , p2 =380mm, v2= 33L, T2 = 273+273=546k
now,
(P1.V1) / T1 = (P2.V2) / T2 [gas law]
(760 x V1) / 273 = (380 x 33) / 546
V1 = 33 / 4 [BY CALCULATING]
V1 = 8.25 L
hope it works
so mark as brainy answer
C2H4 + 3O2 → 2H2O + 2CO2
3 mols of O2 are required to fully combust 1 mol of C2H4
11 litres required 3/1 * 11=33 litre's of oxygen Now , p1= 760mm, T1=273k , v1=? , p2 =380mm, v2= 33L, T2 = 273+273=546k
now,
(P1.V1) / T1 = (P2.V2) / T2 [gas law]
(760 x V1) / 273 = (380 x 33) / 546
V1 = 33 / 4 [BY CALCULATING]
V1 = 8.25 L
hope it works
so mark as brainy answer
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