Đáp án:
\(\begin{array}{l}
b)\\
{m_{Fe}} = 8,4g\\
{m_{Cu}} = 1,6g\\
c)\\
{C_{{M_{HCl}}}} = 1,7M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,15mol\\
{m_{Fe}} = n \times M = 0,15 \times 56 = 8,4g\\
{m_{Cu}} = 10 - 8,4 = 1,6g\\
c)\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{{1,6}}{{80}} = 0,02mol\\
{n_{HCl}} = 2{n_{Fe}} + 2{n_{CuO}} = 0,34mol\\
{C_{{M_{HCl}}}} = \dfrac{n}{V} = \dfrac{{0,34}}{{0,2}} = 1,7M
\end{array}\)