Đáp án:
\(\begin{array}{l}
6)\\
{m_{N{a_2}S{O_4}}} = 14,2g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
6)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{{10}}{{40}} = 0,25\,mol\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,1 \times 1 = 0,2\,mol\\
\dfrac{{0,25}}{2} > \dfrac{{0,1}}{1} \Rightarrow\text{ NaOH dư}\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1\,mol\\
{m_{N{a_2}S{O_4}}} = n \times M = 0,1 \times 142 = 14,2g\\
\text{ Quỳ tím hóa xanh vì NaOH dư}\\
7)\\
FeO + 2HCl \to FeC{l_2} + {H_2}O\\
F{e_2}{O_3} + 6HCl \to 2FeC{l_3} + 3{H_2}O\\
F{e_3}{O_4} + 8HCl \to 2FeC{l_3} + FeC{l_2} + 4{H_2}O
\end{array}\)
