Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
\(\begin{array}{l}
N{a_2}O + {H_2}O \to 2NaOH\\
2NaOH + CuS{O_4} \to N{a_2}S{O_4} + Cu{(OH)_2}\\
Cu{(OH)_2} \to CuO + {H_2}O
\end{array}\)
A: NaOH
B: \(Cu{(OH)_2}\)
C: \(N{a_2}S{O_4}\) và \(CuS{O_4}\) dư
D: CuO
\(\begin{array}{l}
{n_{N{a_2}O}} = 0,1mol\\
\to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,2mol\\
{n_{CuS{O_4}}} = \dfrac{{200 \times 16}}{{100 \times 160}} = 0,2mol\\
\to \dfrac{{{n_{NaOH}}}}{2} < {n_{CuS{O_4}}}
\end{array}\)
Suy ra \(CuS{O_4}\) dư
\(\begin{array}{l}
\to {n_{Cu{{(OH)}_2}}} = \dfrac{1}{2}{n_{NaOH}} = 0,1mol\\
\to {m_B} = {m_{Cu{{(OH)}_2}}} = 9,8g\\
{n_{CuS{O_4}}}dư= 0,2 - \dfrac{1}{2}{n_{NaOH}} = 0,1mol\\
\to {m_{CuS{O_4}}}dư= 16g\\
{n_{N{a_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,1mol\\
\to {m_{N{a_2}S{O_4}}} = 14,2g\\
{m_{dd}} = {m_{N{a_2}O}} + {m_{{H_2}O}} + {m_{CuS{O_4}}}dd - {m_{Cu{{(OH)}_2}}} = 390,2g\\
\to C{\% _{N{a_2}S{O_4}}} = \dfrac{{14,2}}{{390,2}} \times 100\% = 3,64\% \\
\to C{\% _{CuS{O_4}}} = \dfrac{{16}}{{390,2}} \times 100\% = 4,1\%
\end{array}\)
\(\begin{array}{l}
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,1mol\\
\to {n_{HCl}} = 2{n_{CuO}} = 0,2mol\\
\to {V_{HCl}} = \dfrac{{0,2}}{2} = 0,1l
\end{array}\)
