Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
9,
\(\begin{array}{l}
2NaOH + C{O_2} \to N{a_2}C{O_3} + {H_2}O\\
{n_{NaOH}} = 0,16mol\\
{n_{C{O_2}}} = 0,07mol\\
\dfrac{{{n_{NaOH}}}}{2} > {n_{C{O_2}}}
\end{array}\)
Suy ra có NaOH dư
Vậy NaOH dư
\(\begin{array}{l}
\to {n_{NaOH}}dư= 0,16 - 2{n_{C{O_2}}} = 0,02mol\\
\to {m_{NaOH}}dư= 0,8g\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,07mol\\
\to {m_{N{a_2}C{O_3}}} = 7,42g
\end{array}\)
10,
a, Nguyên tố dinh dưỡng có trong loại phân tren là N (Nito)
b,
\(\begin{array}{l}
{n_{{{(N{H_4})}_2}S{O_4}}} = 3,78mol\\
\to {n_N} = 2{n_{{{(N{H_4})}_2}S{O_4}}} = 7,56mol\\
\to {m_N} = 105,84g\\
\to \% {m_N} = \dfrac{{105,84}}{{500}} \times 100\% = 21,168\%
\end{array}\)
c,
\(\begin{array}{l}
{n_{{{(N{H_4})}_2}S{O_4}}} = 3,78mol\\
\to {n_N} = 2{n_{{{(N{H_4})}_2}S{O_4}}} = 7,56mol\\
\to {m_N} = 105,84g
\end{array}\)
11,
\(\begin{array}{l}
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{CuC{l_2}}} = 0,2mol\\
{n_{NaOH}} = 0,5mol\\
{n_{CuC{l_2}}} < \dfrac{{{n_{NaOH}}}}{2}
\end{array}\)
Suy ra có NaOH dư sau phản ứng
\(\begin{array}{l}
\to {n_{Cu{{(OH)}_2}}} = {n_{CuC{l_2}}} = 0,2mol\\
\to {n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,2mol\\
\to {m_{CuO}} = 16g
\end{array}\)
Chất tan trong nước lọc là: NaOH dư và NaCl
\(\begin{array}{l}
\to {n_{NaOH}}dư= 0,5 - 2{n_{CuC{l_2}}} = 0,1mol\\
\to {m_{NaOH}}dư= 4g\\
{n_{NaCl}} = 2{n_{CuC{l_2}}} = 0,4mol\\
\to {m_{NaCl}} = 23,4g\\
\to m = {m_{NaOH}}dư+ {m_{NaCl}} = 27,4g
\end{array}\)
12,
\(\begin{array}{l}
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,02mol\\
\to {n_{NaOH}} = 2{n_{{H_2}S{O_4}}} = 0,04mol\\
\to {m_{NaOH}} = 1,6g\\
\to {m_{NaOH}}{\rm{dd}} = \dfrac{{1,6 \times 100}}{{20}} = 8g\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,02mol\\
\to {n_{KOH}} = 2{n_{{H_2}S{O_4}}} = 0,04mol\\
\to {m_{KOH}} = 2,24g\\
\to {m_{KOH}}{\rm{dd}} = \dfrac{{2,24 \times 100}}{{5,6}} = 40g\\
\to {V_{KOH}}{\rm{dd}} = \dfrac{{40}}{{1,045}} = 38,28ml
\end{array}\)
