ta có Mg+H2SO4=>MgSO4+H2
a,ta có mMg=12(g)=>nMg=$\frac{12}{24}$=0.5(mol)
=>nH2=0.5(mol)=>vH2=0.5*22.4=11.2(lít)
b,nH2SO4=0.5(mol)=>mH2SO4=0.5*(2+32+16*4)=49(g)
nMgSO4=0.5(mol)=>mMgSO4=0.5*(24+32+16*4)=60(g)
b,Fe2O3+3H2=>2Fe+3H2O
mFe2O3=16(g)=>nFe2O3=$\frac{16}{56*2+16*3}$=0.1(mol)
ta có tỉ lệ:nFe2O3:nH2=$\frac{0.1}{1}$<$\frac{0.5}{3}$(Fe2O3 hết,H2 dư)
=>nFe=0.1*2=0.2(mol)
=>mFe=0.2*56=11.2(g)