Đáp án:

a, \({m_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{29,4 \times 100}}{{10}} = 294g\)

b, \(C{\% _{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{0,1 \times 400}}{{310}} \times 100\%  = 12,9\% \)

Giải thích các bước giải:

\(\begin{array}{l}
F{e_2}{O_3} + 3{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{F{e_2}{O_3}}} = 0,1mol\\
 \to {n_{{H_2}S{O_4}}} = 3{n_{F{e_2}{O_3}}} = 0,3mol\\
 \to {m_{{H_2}S{O_4}}} = 29,4g\\
 \to {m_{{H_2}S{O_4}}}{\rm{dd}} = \dfrac{{29,4 \times 100}}{{10}} = 294g\\
{m_{{\rm{dd}}}} = {m_{F{e_2}{O_3}}} + {m_{{H_2}S{O_4}}}{\rm{dd}} = 16 + 294 = 310g\\
 \to {n_{F{e_2}{{(S{O_4})}_3}}} = {n_{F{e_2}{O_3}}} = 0,1mol\\
 \to C{\% _{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{0,1 \times 400}}{{310}} \times 100\%  = 12,9\% 
\end{array}\)

Đáp án + Giải thích các bước giải:

\begin{array}{l}n_{F{e_2}{O_3}}=\dfrac{16}{160}=0.1(mol)\\PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2\\\text{Theo pt:}n_{H_2SO_4}=3n_{Fe_2O_3}=3\times0.1=0.3(mol)\\m_{ddH_2SO_4}=\dfrac{0.3\times98\times100%}{10%}=294(g)\\m_{\text{dd spư}}=16+294=310(g)\\\text{Theo pt:}n_{Fe_2(SO_4)_3}=n_{Fe_2O_3}=0.1(mol)\\C\%_{ddFe_2(SO_4)_3}=\dfrac{0.1\times400}{310}\times100\%=12.9\%\end{array}