Đáp án:
\(\begin{array}{l}
a)\\
{m_{Mg}} = 3,6g\\
b)\\
{V_{{H_2}}} = 3,36l\\
c)\\
{m_{Fe}} = 5,04g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{MgC{l_2}}} = \dfrac{{14,25}}{{95}} = 0,15\,mol\\
{n_{Mg}} = {n_{MgC{l_2}}} = 0,15\,mol\\
{m_{Mg}} = 0,15 \times 24 = 3,6g\\
b)\\
{n_{{H_2}}} = {n_{Mg}} = 0,15\,mol\\
{V_{{H_2}}} = 0,15 \times 22,4 = 3,36l\\
c)\\
F{e_2}{O_3} + 3{H_2} \to 2Fe + 3{H_2}O\\
{n_{Fe}} = 0,15 \times \frac{2}{3} = 0,1\,mol\\
{m_{Fe}} = 0,1 \times 56 \times 90\% = 5,04g
\end{array}\)