Đáp án:
\(C{\% _{CaS{O_4}}} = 33,5\% \)
Giải thích các bước giải:
\(\begin{array}{l}
Ca + {H_2}S{O_4} \to CaS{O_4} + {H_2}\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{n_{Ca}} = {n_{{H_2}}} = 0,4mol\\
{m_{Ca}} = n \times M = 0,4 \times 40 = 16g\\
{m_{{\rm{dd}}spu}} = {m_{Ca}} + {m_{{H_2}S{O_4}}} - {m_{{H_2}}} = 16 + 147 - 0,4 \times 2 = 162,2g\\
{n_{CaS{O_4}}} = {n_{{H_2}}} = 0,4mol\\
{m_{CaS{O_4}}} = n \times M = 0,4 \times 136 = 54,4g\\
C{\% _{CaS{O_4}}} = \dfrac{{54,4}}{{162,2}} \times 100\% = 33,5\%
\end{array}\)
