Đáp án:

 a) mFe=7 gam

b) CM HCl=3,125M

c) mCu=8 gam

Giải thích các bước giải:

Phản ứng xảy ra:

\(Fe + 2HCl\xrightarrow{{}}FeC{l_2} + {H_2}\)

Ta có: \({n_{{H_2}}} = \frac{{2,8}}{{22,4}} = 0,125{\text{ mol = }}{{\text{n}}_{Fe}} \to {m_{Fe}} = 0,125.56 = 7{\text{ gam}}\)

\({n_{HCl}} = 2{n_{{H_2}}} = 0,125.2 = 0,25{\text{ mol}} \to {{\text{C}}_{M{\text{ HCl}}}} = \frac{{0,25}}{{0,08}} = 3,125{\text{M}}\)

Dẫn H2 qua CuO nung nóng

\(CuO + {H_2}\xrightarrow{{}}Cu + {H_2}O\)

\({n_{CuO}} = \frac{{12,8}}{{64 + 16}} = 0,16{\text{ mol > }}{{\text{n}}_{{H_2}}}\)

Vậy nên CuO dư.

\(\to {n_{Cu}} = {n_{{H_2}}} = 0,125{\text{ mol}} \to {{\text{m}}_{Cu}} = 0,125.64 = 8{\text{ gam}}\)