Đáp án:
a) 11,65g
b) 100g
c) 4,88%
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuS{O_4} + BaC{l_2} \to CuC{l_2} + BaS{O_4}\\
{n_{CuS{O_4}}} = \dfrac{{50 \times 16\% }}{{160}} = 0,05\,mol\\
{n_{BaS{O_4}}} = {n_{CuS{O_4}}} = 0,05\,mol\\
{m_{BaS{O_4}}} = 0,05 \times 233 = 11,65g\\
b)\\
{n_{BaC{l_2}}} = {n_{CuS{O_4}}} = 0,05\,mol\\
{m_{{\rm{dd}}BaC{l_2}}} = \dfrac{{0,05 \times 208}}{{10,4\% }} = 100g\\
c)\\
{m_{{\rm{dd}}spu}} = 50 + 100 - 11,65 = 138,35g\\
{n_{CuC{l_2}}} = {n_{CuS{O_4}}} = 0,05\,mol\\
{C_\% }CuC{l_2} = \dfrac{{0,05 \times 135}}{{138,35}} \times 100\% = 4,88\%
\end{array}\)
