Đáp án:
17,C
4,
a, \({m_{Na}} = 11,5g\) và \({m_{Fe}} = 8,5g\)
b, \(C\% NaOH = \)9,11% và \(C{M_{NaOH}} = \dfrac{{0,5}}{{0,2}} = 2,5M\)
Giải thích các bước giải:
17, \({S_{KN{O_3}}} = \dfrac{{40}}{{95}} \times 100 = 42,11\)
3,
\(\begin{array}{l}
{K_2}O + {H_2}O \to 2KOH\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Mg + \dfrac{1}{2}{O_2} \to MgO\\
S{O_2} + {H_2}O \to {H_2}S{O_3}\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
FeO + {H_2}S{O_4} \to FeS{O_4} + {H_2}O
\end{array}\)
4,
\(\begin{array}{l}
Na + {H_2}O \to NaOH + \frac{1}{2}{H_2}\\
{n_{{H_2}}} = 0,25mol\\
\to {n_{Na}} = 2{n_{{H_2}}} = 0,5mol\\
\to {m_{Na}} = 11,5g\\
\to {m_{Fe}} = 20 - 11,5 = 8,5g\\
{n_{NaOH}} = {n_{Na}} = 0,5mol\\
{m_{{H_2}O}} = 1 \times 200 = 200g\\
\to {m_{{\rm{dd}}}} = 20 + 200 - 0,25 \times 2 = 219,5g\\
\to C\% NaOH = \dfrac{{0,5 \times 40}}{{219,5}} \times 100\% = 9,11\% \\
\to C{M_{NaOH}} = \dfrac{{0,5}}{{0,2}} = 2,5M
\end{array}\)
