A gaseous alkane requires two times its volume for complete combustion
molecular formula of the compound is
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Hii There!!!!
General equation for the complete combustion of alkane is
CnH2n+2 + (3n+1)/2 O₂ = nCO₂ + (n+1)H₂O
here the co-efficients of the equation represent the relative mole no.
for gases at NTP 1 mole occupies 22.4 litre
so relative mole no can b considered as relative volume
here (3n+1)/2 = 5
so n = 3
=) alkane is C₃H₈.
Hope it helps
General equation for the complete combustion of alkane is
CnH2n+2 + (3n+1)/2 O₂ = nCO₂ + (n+1)H₂O
here the co-efficients of the equation represent the relative mole no.
for gases at NTP 1 mole occupies 22.4 litre
so relative mole no can b considered as relative volume
here (3n+1)/2 = 5
so n = 3
=) alkane is C₃H₈.
Hope it helps
Answered by
2
General equation for the complete combustion of alkane is
CnH2n+2 + (3n+1)/2 O₂ = nCO₂ + (n+1)H₂O
here the co-efficients of the equation represent the relative mole no.
for gases at NTP 1 mole occupies 22.4 litre
so relative mole no can b considered as relative volume
here (3n+1)/2 = 5
so n = 3
& the alkane is C₃H₈
Hope This Helps
CnH2n+2 + (3n+1)/2 O₂ = nCO₂ + (n+1)H₂O
here the co-efficients of the equation represent the relative mole no.
for gases at NTP 1 mole occupies 22.4 litre
so relative mole no can b considered as relative volume
here (3n+1)/2 = 5
so n = 3
& the alkane is C₃H₈
Hope This Helps
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