Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Mg}} = 20\% \\
\% {m_{FeO}} = 80\% \\
b)\\
{m_{FeC{l_2}}} = 25,4g\\
{m_{MgC{l_2}}} = 14,25g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,15\,mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
{m_{FeO}} = 18 - 3,6 = 14,4g\\
\% {m_{Mg}} = \dfrac{{3,6}}{{18}} \times 100\% = 20\% \\
\% {m_{FeO}} = 100 - 20 = 80\% \\
b)\\
{n_{FeO}} = \dfrac{{14,4}}{{72}} = 0,2\,mol\\
{n_{FeC{l_2}}} = {n_{FeO}} = 0,2\,mol\\
{n_{MgC{l_2}}} = {n_{Mg}} = 0,15\,mol\\
{m_{FeC{l_2}}} = 0,2 \times 127 = 25,4g\\
{m_{MgC{l_2}}} = 0,15 \times 95 = 14,25g
\end{array}\)