Đáp án:

 b) 11,2 g và 12,8 g

c) 43,24% và 56,76%

d) e) 9,184l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Fe + CuS{O_4} \to FeS{O_4} + Cu\\
b)\\
nFe\,pu = a\,mol\\
64a - 56a = 29,6 - 28\\
 =  > a = 0,2mol\\
mFepu = 0,2 \times 56 = 11,2g\\
mCu = 0,2 \times 64 = 12,8g\\
c)\\
\% mCu = \dfrac{{12,8}}{{29,6}} \times 100\%  = 43,24\% \\
\% mFe = 100 - 43,24 = 56,76\% \\
d)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
nFe = \dfrac{{23}}{{56}} = 0,41\,mol\\
n{H_2} = nFe = 0,41\,mol\\
V{H_2} = 0,41 \times 22,4 = 9,184l\\
e)\\
Fe + 4HN{O_3} \to Fe{(N{O_3})_3} + NO + 2{H_2}O\\
nNO = nFe = 0,41\,mol\\
VNO = 0,41 \times 22,4 = 9,184l
\end{array}\)

\(\begin{array}{l}
a)Fe+CuSO_4\to FeSO_4+Cu\\
b)\\
Tang\,giam\,khoi\,luong:\\
n_{Fe}=n_{Cu}=\frac{29,6-28}{64-56}=0,2(mol)\\
\to m_{Fe}=0,2.56=11,2(g);m_{Cu}=0,2.64=12,8(g)\\
c)\\
\%m_{Fe}=\frac{28-11,2}{29,6}.100\%\approx 56,76\%\\
\%m_{Cu}=100-56,76=43,24\%\\
d)\\
Fe+2HCl\to FeCl_2+H_2\\
Theo\,PT:\,n_{H_2}=n_{Fe}=\frac{23}{56}(mol)\\
\to V_{H_2}=\frac{23}{56}.22,4=9,2(l)\\
e)\\
Fe+4HNO_3\to Fe(NO_3)_3+NO+2H_2O\\
Theo\,PT:\,n_{NO}=n_{Fe}=\frac{23}{56}(mol)\\
\to V_{NO}=\frac{23}{56}.22,4=9,2(l)
\end{array}\)