Đáp án:
\(\begin{array}{l}
C{\% _{{\rm{CuS}}{{\rm{O}}_4}}} = 14,81\% \\
C{\% _{{H_2}S{O_4}(dư)}} = 9,07\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)CuO + {H_2}S{O_4} \to C{\rm{uS}}{O_4} + {H_2}O\\
b)\\
{n_{CuO}} = 0,1mol\\
{m_{{H_2}S{O_4}}} = \dfrac{{100 \times 19,6\% }}{{100\% }} = 19,6g\\
\to {n_{{H_2}S{O_4}}} = 0,2mol\\
\to {n_{{H_2}S{O_4}}} > {n_{CuO}} \to {n_{{H_2}S{O_4}}}dư\\
\to {n_{{H_2}S{O_4}}} = {n_{CuO}} = 0,1mol\\
\to {n_{{H_2}S{O_4}(dư)}} = 0,1mol \to {m_{{H_2}S{O_4}(dư)}} = 9,8g\\
{n_{{\rm{CuS}}{{\rm{O}}_4}}} = {n_{CuO}} = 0,1mol \to {m_{{\rm{CuS}}{{\rm{O}}_4}}} = 16g\\
{m_{{\rm{dd}}}} = {m_{CuO}} + {m_{{\rm{dd}}{H_2}S{O_4}}} = 108g\\
\to C{\% _{{\rm{CuS}}{{\rm{O}}_4}}} = \dfrac{{16}}{{108}} \times 100\% = 14,81\%
\end{array}\)
\( \to C{\% _{{H_2}S{O_4}(dư)}} = \dfrac{{9,8}}{{108}} \times 100\% = 9,07\% \)
