Đáp án:
\(\begin{array}{l}
a)\\
{C_\% }HCl = 48,67\% \\
b)\\
{m_{MgC{l_2}}} = 95g\\
c)\\
{V_{{\rm{dd}}NaOH}} = 1l\\
d)\\
{m_{Fe}} = 42g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{24}}{{24}} = 1\,mol\\
{n_{HCl}} = 2{n_{Mg}} = 2\,mol\\
{C_\% }HCl = \dfrac{{2 \times 36,5}}{{150}} \times 100\% = 48,67\% \\
b)\\
{n_{MgC{l_2}}} = {n_{Mg}} = 1\,mol\\
{m_{MgC{l_2}}} = 1 \times 95 = 95g\\
c)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}} = 2\,mol\\
{V_{{\rm{dd}}NaOH}} = \dfrac{2}{2} = 1l\\
d)\\
F{e_3}{O_4} + 4{H_2} \xrightarrow{t^0} 3Fe + 4{H_2}O\\
{n_{{H_2}}} = {n_{Mg}} = 1\,mol\\
{n_{Fe}} = 1 \times \dfrac{3}{4} = 0,75\,mol\\
{m_{Fe}} = 0,75 \times 56 = 42g
\end{array}\)
