Bài 4:
nC3H8= $\frac{17,6}{44}$= 0,4 mol
C3H8+ 5O2 $\buildrel{{t^o}}\over\longrightarrow$ 3CO2+ 4H2O
=> nCO2= 1,2 mol; nH2O= 1,6 mol
a, V CO2= 1,2.22,4= 26,88l
b, mH2O= 1,6.18= 28,8g
Bài 5:
nO2= 1,5 mol
a, C+ O2 $\buildrel{{t^o}}\over\longrightarrow$
=> nC= 1,5 mol
=> mC= 1,5.12= 18g
b, 2H2+ O2 $\buildrel{{t^o}}\over\longrightarrow$ 2H2O
=> nH2= 3 mol
=> mH2= 3.2= 6g
c, S+ O2 $\buildrel{{t^o}}\over\longrightarrow$ SO2
=> nS= 1,5 mol
=> mS= 1,5.32= 48g
d, 4P+ 5O2 $\buildrel{{t^o}}\over\longrightarrow$ 2P2O5
=> nP= 1,2 mol
=> mP= 1,2.31= 37,2g
