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