$n_{Al}=5,4/27=0,2mol$

$PTHH :$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$

$\text{ a/Theo pt :}$

$n_{H_2}=3/2.n_{Al}=3/2.0,2=0,3mol$

$⇒V_{H_2}=0,3.22,4=6,72l$

$b/PTHH :$

$CuO+H_2\overset{t^o}\to Cu+H_2O(1)$

$Fe_2O_3+3H_2\overset{t^o}\to 2Fe+3H_2O(2)$

$\text{Theo pt (1) :}$

$n_{Cu}=6,4/64=0,1mol$

$⇒n_{CuO}=n_{Cu}=0,1mol$

$⇒m_{CuO}=0,1.80=8g$

$\text{Ta có :}$

$n_{H_2(2)}=n_{H_2}-n_{H_2(1)}=0,3-0,1=0,2mol$

$\text{Theo pt (2) :}$

$n_{Fe_2O_3}=1/3.n_{H_2}=1/3.0,2=\dfrac{0,2}{3}mol$

$⇒m_{Fe_2O_3}=\dfrac{0,2}{3}.160=\dfrac{32}{3}g$

ta có phương trình:

2Al+3H2SO4=>Al2(SO4)3+3H2

a,ta có mAl=5.4(g)=>nAl=$\frac{5.4}{27}$=0.2(mol)

=>nH2=$\frac{0.2*3}{2}$=0.3(mol) 

=>vH2=0.3*22.4=6.72(mol)

b,CuO+H2=>Cu+H2O

 0.1         0.1         0.1          (mol)

Fe2O3+3H2=>2Fe+3H2O

              0.2                          (mol)

 ta có mCu=6.4(g)=>nCu=$\frac{6.4}{64}$=0.1(mol) 

=>nCuO=0.1(mol)=>mCuO=0.1*(64+16)=8(g)

ta có nH2(trong pt CuO)=0.1(mol)

=>nH2(trong Fe2O3)=0.3-0.1=0.2(mol)

=>nFe2O3=$\frac{0.2}{3}$=0.066(mol) 

=>mFe2O3=0.066*(56*2+16*3)=10.56(g)

c,thiếu đề....