Đáp án:
710,8g
4,28%
Giải thích các bước giải:
\(\begin{array}{l}
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
nFe = \dfrac{m}{M} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
m{H_2}S{O_4} = m{\rm{dd}} \times C\% = 700 \times 35\% = 245g\\
n{H_2}S{O_4} = \dfrac{m}{M} = \dfrac{{245}}{{98}} = 2,5\,mol\\
\dfrac{{0,2}}{1} < \dfrac{{2,5}}{1} \Rightarrow\text{$H_2SO_4$ dư} \\
n{H_2} = nFeS{O_4} = nFe = 0,2\,mol\\
m{\rm{dd}}\,spu = 11,2 + 700 - 0,2 \times 2 = 710,8g\\
C\% FeS{O_4} = \dfrac{{mct}}{{m{\rm{dd}}}} \times 100\% = \dfrac{{0,2 \times 152}}{{710,8}} \times 100\% = 4,28\%
\end{array}\)