Đáp án:
\(\begin{array}{l}
a)\\
{m_{CuO}} = 16g\\
{m_{Cu}} = 6,4g\\
{C_\% }HCl = 7,3\% \\
b)\\
{C_\% }CuS{O_4} = 16\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuC{l_2}}} = \dfrac{{27}}{{135}} = 0,2\,mol\\
{n_{CuO}} = {n_{CuC{l_2}}} = 0,2\,mol\\
{m_{CuO}} = 0,2 \times 80 = 16g\\
{m_{Cu}} = 22,4 - 16 = 6,4g\\
{n_{HCl}} = 2{n_{CuC{l_2}}} = 0,4\,mol\\
{C_\% }HCl = \dfrac{{0,4 \times 36,5}}{{200}} \times 100\% = 7,3\% \\
b)\\
Cu + 2{H_2}S{O_4} \to CuS{O_4} + S{O_2} + 2{H_2}O\\
{n_{S{O_2}}} = {n_{CuS{O_4}}} = {n_{Cu}} = \dfrac{{6,4}}{{64}} = 0,1\,mol\\
{m_{{\rm{dd}}spu}} = 6,4 + 100 - 0,1 \times 64 = 100g\\
{C_\% }CuS{O_4} = \dfrac{{0,1 \times 160}}{{100}} \times 100\% = 16\%
\end{array}\)