Giải thích các bước giải:

Câu 1: 

$a)$

$n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$

$n_{H_2SO_4}=0,2×2=0,4(mol)$

$Fe+H_2SO_4\to FeSO_4+H_2$

$n_{Fe}<n_{H_2SO_4}\to H_2SO_4$ dư.

$n_{H_2}=n_{Fe}=0,2(mol)$

$V_{H_2}=0,2×22,4=4,48(l)$

$b)$

$n_{H_2SO_4\ \rm (dư)}=0,4-0,2=0,2(g)$

$m_{H_2SO_4\ \rm (dư)}=0,2×98=19,6(g)$

Câu 2:

$a)$

$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

$n_{HCl}=3n_{Al}=0,6(mol)$

$C_{M\ HCl}=\dfrac{0,6}{0,4}=1,5(M)$

$b)$

$n_{CuO}=\dfrac{32}{80}=0,4(mol)$

$n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$n_{CuO}>n_{H_2}\to CuO$ dư.

$n_{CuO\ \rm (dư)}=0,4-0,3=0,1(mol)$

$n_{Cu}=n_{H_2}=0,3(mol)$

$\%m_{CuO\ \rm (dư)}=\dfrac{0,1×80}{0,1×80+0,3×64}×100\%=29,41\%$

$\%m_{Cu}=100\%-29,41\%=70,59\%$

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