câu 1:
a,ta có phương trình:4Al+3O2=>2Al2O3
b,ta có:mAl=10.8(g)=>nAl=$\frac{10.8}{27}$=0.4(mol)
=>nAl2O3=$\frac{0.4*2}{4}$=0.2(mol)
=>mAl2O3=0.2*(27*2+16*3)=20.4(g)
c,nO2=$\frac{0.4*3}{4}$=0.3(mol)
=>vO2=0.3*22.4=6.72(lít)
c,KClO3=>KCl+$\frac{3}{2}$O2
ta có nO2=0.3(mol)
=>nKClO3=$\frac{0.3*2}{3}$=0.2(mol)
=>mKClO3=0.2*(39+35.5+16*3)=24.5(g)
câu 2:
a,2Al+3H2SO4=>Al2(SO4)3+3H2
b,vH2=6.72(lít)=>nH2=6.72/22.4=0.3(mol)
=>nAl2(SO4)3=$\frac{0.3}{3}$=0.1(mol)
=>mAl2(SO4)3=0.1*(27*2+32*3+16*12)=34.2(g)
c,2H2O=>2H2+O2
nH2=0.3(mol)
=>nH2O=nH2=0.3(mol)
=>mH2O=0.3*18=5.4(g)
câu 3:
a,Fe3O4+4H2=>3Fe+4H2O
b,ta có mFe3O4=46.4(g)=>nFe3O4=$\frac{46.4}{56*3+16*4}$=0.2(mol)
vH2=8.96(lít)=>nH2=$\frac{8.96}{22.4}$=0.4(mol)
ta có tỉ lệ:nFe3O4:nH2=$\frac{0.2}{1}$>$\frac{0.4}{4}$(Fe3O4 dư,H2 hết)
=>nFe=$\frac{0.4*3}{4}$=0.3(mol) =>mFe=0.3*56=16.8(g)
nH2O=nH2=0.4(mol)=>mH2O=0.4*18=7.2(g)