Đáp án:

nFe=28/56=0.5(mol)

a. Fe+H₂SO₄→H₂+FeSO₄

b. Theo PT ta có:

nH2=nFe=0.5(mol)

=>vH2=0.5*22.4=11.2(lít)

c. PTHH

H2+CuO->Cu+H2O

nCuO=12/80=0.15(mol)

Xét: 0.5/1>0.15/1=>nCuO hết, nH2 dư

=> nH2=nCuO=0.15(mol)

=>nH2 dư=0.5-0.15=0.35(mol)

=>mH2 dư=0.35x2=0.7(g)

Em tham khảo nha :

\(\begin{array}{l}
a)\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{{28}}{{56}} = 0,5mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,5mol\\
{V_{{H_2}}} = 0,5 \times 22,4 = 11,2l\\
c)\\
{n_{CuO}} = \dfrac{{12}}{{80}} = 0,15mol\\
CuO + {H_2} \to Cu + {H_2}O\\
\dfrac{{0,15}}{1} < \dfrac{{0,5}}{1} \Rightarrow {H_2}\text{ dư}\\
{n_{{H_2}d}} = {n_{{H_2}}} - {n_{CuO}} = 0,5 - 0,15 = 0,35mol\\
{m_{{H_2}d}} = 0,35 \times 2 = 0,7g
\end{array}\)