Đáp án:
\(\begin{array}{l}
b)\\
{m_{Fe}} = 25,2g\\
c)\\
{m_{NaCl}} = 52,65g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{10,08}}{{22,4}} = 0,45mol\\
{n_{Fe}} = {n_{{H_2}}} = 0,45mol\\
{m_{Fe}} = n \times M = 0,45 \times 56 = 25,2g\\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,9mol\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaCl}} = {n_{HCl}} = 0,9mol\\
{m_{NaCl}} = n \times M = 0,9 \times 58,5 = 52,65g
\end{array}\)