hihi chào cậu
PTHH:\(C4H10+\dfrac{13}{2}O2\rightarrow4CO2+5H2O\)
mk cân bằn luôn
b, ta có C4H10=26.4%propan + 69.6%buttan+ 4%ko cháy
ta có: 1.26.4:100=0.264(g)
=1.69.6:100=0.696(g)
=1-(0.264+0.696)=0.04
PTHH:C4H10+13/2O2----->4CO2+5H2O
mol: 1 7.5 4 5
vậy nO2=13/2
VO2(đktc)=13/2.22.4=145.6(l)