Đáp án:
\(\begin{array}{l}
b)\\
{C_{{M_{NaOH}}}} = 0,8M\\
c)\\
{m_{Cu{{(OH)}_2}}} = 19,6g\\
d)\\
{V_{{H_2}S{O_4}}} = 85,96ml
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
b)\\
{n_{N{a_2}O}} = \dfrac{m}{M} = \dfrac{{12,4}}{{62}} = 0,2mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,4mol\\
{C_{{M_{NaOH}}}} = \dfrac{n}{V} = \dfrac{{0,4}}{{0,5}} = 0,8M\\
c)\\
2NaOH + CuC{l_2} \to 2NaCl + Cu{(OH)_2}\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{NaOH}}}}{2} = \dfrac{{0,4}}{2} = 0,2mol\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,2 \times 98 = 19,6g\\
d)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,2mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,2 \times 98 = 19,6g\\
{m_{dd{H_2}S{O_4}}} = \dfrac{{19,6 \times 100}}{{20}} = 98g\\
{V_{{H_2}S{O_4}}} = \dfrac{m}{d} = \dfrac{{98}}{{1,14}} = 85,96ml
\end{array}\)
