Bài 1.
mS=$\frac{36×0.5}{100}$=0.18(g)
m(tạp chất)= $\frac{36×1.5}{100}$ =0.54(g)
⇒mC=36-0.18-0.54=35.28(g)
nC=$\frac{35.28}{12}$ =2.94(mol)
nS=$\frac{0.18}{32}$=0.005625(mol)
C+O2⇒CO2
2.94.............2.94
S+O2⇒SO2
0.005625...........0.005625
⇒VCO2(đktc)=2.94×22.4=65.856(l)
VSO2(đktc)=0.005625×22.4=0.126(l)
Bài 2.
1. 2KMnO4⇒MnO2+O2+K2MnO4
2.nKMnO4=$\frac{63.2}{158}$=0.4(mol)
nO2=$\frac{1}{2}$ nKMnO4=0.2(mol)
⇒VO2=0.2×22.4=4.48(l)
3.
3Fe+2O2⇒Fe3O4
0.2......0.1
nFe=22.4÷56=0.4(mol)
vì nO2<nFe(0.2<0.4)⇒Fe dư, O2 hết
⇒mFe3O4=0.1×(56×3+16×4)=23.2(g)
Bài 3.
2C3H6+9O2⇒6CO2+6H2O
V(tạp chất)=$\frac{5×2}{100}$ =0.1(l)
⇒VC3H6(cháy)=2-0.1=0.9(mol)
nC3H6=0.9÷22.4=0.04(mol)
mà nO2=$\frac{9}{2}$ nC3H6=0.18(mol)
⇒VO2=0.18×22.4=4.032(mol)
⇒V(không khí)=$\frac{4.032×80}{20}$ =16.128(l)
