Đáp án:
b)
\(\begin{array}{l}
{m_{Cu{{(OH)}_2}}} = 9,8g\\
{m_{CuO}} = 8g\\
c)\\
{C_{{M_{NaOH}}}} = 1M\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{1}{3}M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
CuS{O_4} + 2NaOH \to N{a_2}S{O_4} + Cu{(OH)_2}\\
b)\\
{n_{CuS{O_4}}} = {C_M} \times V = 0,1 \times 1 = 0,1mol\\
{n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,1mol\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,1 \times 98 = 9,8g\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,1mol\\
{m_{CuO}} = n \times M = 0,1 \times 80 = 8g\\
c)\\
{n_{NaOH}} = 2{n_{CuS{O_4}}} = 2 \times 0,1 = 0,2mol\\
{C_{{M_{NaOH}}}} = \dfrac{n}{V} = \dfrac{{0,2}}{{0,2}} = 1M\\
{n_{N{a_2}S{O_4}}} = {n_{CuS{O_4}}} = 0,1mol\\
{C_{{M_{N{a_2}S{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,1 + 0,2}} = \dfrac{1}{3}M
\end{array}\)