Đáp án:
Bạn tham khảo lời giải ở dưới nhé !!!
Giải thích các bước giải:
\(\begin{array}{l}
11.\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
a)\\
{n_{{H_2}}} = 0,2mol\\
\to {n_{Zn}} = {n_{{H_2}}} = 0,2mol\\
\to {m_{Zn}} = 13g\\
b)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,4mol\\
\to C{M_{HCl}} = \dfrac{{0,4}}{{0,1}} = 4M\\
12.\\
ZnO + {H_2}S{O_4} \to ZnS{O_4} + {H_2}O\\
a)\\
{n_{ZnO}} = 0,1mol\\
\to {n_{ZnS{O_4}}} = {n_{ZnO}} = 0,1mol\\
\to {m_{ZnS{O_4}}} = 16,1g\\
b)\\
{n_{{H_2}S{O_4}}} = {n_{ZnO}} = 0,1mol\\
\to {V_{{H_2}S{O_4}}} = \dfrac{{0,1}}{4} = 0,025l\\
13.\\
N{a_2}O + {H_2}O \to 2NaOH\\
a)\\
{n_{N{a_2}O}} = 0,25mol\\
\to {n_{NaOH}} = 2{n_{N{a_2}O}} = 0,5mol\\
\to C{M_{NaOH}} = \dfrac{{0,5}}{{0,5}} = 1M\\
b)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{NaOH}} = 0,25mol\\
\to {m_{{H_2}S{O_4}}} = 24,5g\\
\to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{24,5}}{{20\% }} \times 100\% = 122,5g
\end{array}\)
