câu 3:
CuO+H2=>Cu+H2O
a,ta có mCuO=40(g)=>nCuO=$\frac{40}{64+16}$=0.5(mol)
ta có nH2=nCu=nCuO=0.5(mol)
=>vH2=0.5*22.4=11.2(lít)
mCu=0.5*64=32(g)
b,Mg+2HCl=>MgCl2+h2
nh2=0.5(mol)
ta có nMg=nH2=0.5(mol)
=>mMg=0.5*24=12(g)
nHCl=0.5*2=1(mol)
=>mHCl=1*36.5=36.5(g)
bài 5:
Fe2O3+3H2=>2Fe+3H2O
a,ta có mFe2O3=40(g)=>nFe2O3=$\frac{40}{56*2+16*3}$=0.25(mol)
=>nH2=0.25*3=0.75(mol)
=>vH2=0.75*22.4=16.8(lít)
b,nFe=0.25*2=0.5(mol)
=>mFe=0.5*56=28(g)
c,2Al+6HCl=>2AlCl3+3H2
nH2=0.75(mol)
=>nAl=$\frac{0.75*2}{3}$=0.5(mol)
=>mAl=0.5*27=13.5(g)
nHCl=$\frac{0.75*6}{3}$=1.5(mol)
=>mHCl=1.5*36.5=54.75(g)
