Đáp án:

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Giải thích các bước giải:

\(\begin{array}{l}
K + HCl \to KCl + \frac{1}{2}{H_2}\\
{K_2}O + 2HCl \to 2KCl + {H_2}O\\
KCl + AgN{O_3} \to AgCl + KN{O_3}\\
{n_{{H_2}}} = 0,05mol\\
 \to {n_{HCl}}(K) = 2{n_{{H_2}}} = 0,1mol\\
 \to {m_{HCl}}(K) = 3,65g\\
{n_{AgCl}} = \dfrac{{10,7625}}{{143,5}} = 0,075mol\\
 \to {n_{AgN{O_3}}} = {n_{AgCl}} = 0,075mol\\
 \to {m_{AgN{O_3}}} = 12,75g\\
 \to {m_{AgN{O_3}}}thực tế= 12,75 + 12,75 \times 10\%  = 14,025g\\
 \to {n_{AgN{O_3}}}thực tế= 0,0825mol\\
 \to {V_2} = \dfrac{{0,0825}}{2} = 0,04125l = 41,25ml\\
{n_K} = 2{n_{{H_2}}} = 0,1mol \to {m_K} = 3,9g\\
{n_{KCl}}(AgN{O_3}) = {n_{AgCl}} = 0,075mol\\
 \to {m_{_{KCl}}}(AgN{O_3}) = 5,5875g\\
{n_{KCl}}(K) = 2{n_{{H_2}}} = 0,1mol\\
 \to {m_{KCl}}(K) = 7,45g\\
 \to {m_Y} = 2{m_{_{KCl}}}(AgN{O_3}) = 11,175g\\
 \to {m_{KCl}}({K_2}O) = 11,175 - 7,45 = 3,725g\\
 \to {n_{KCl}}({K_2}O) = 0,05mol\\
 \to {n_{{K_2}O}} = \dfrac{1}{2}{n_{KCl}}({K_2}O) = 0,025mol\\
 \to {m_{{K_2}O}} = 2,35g\\
 \to m = {m_K} + {m_{{K_2}O}} = 6,25g\\
{n_{HCl}} = {n_K} + 2{n_{{K_2}O}} = 0,15mol\\
 \to {m_{HCldd}} = \dfrac{{0,15 \times 36,5 \times 100}}{{20}} = 27,375g\\
 \to {V_1} = \dfrac{m}{D} = \dfrac{{27,375}}{{1,19}} = 23ml\\
\% {m_K} = \dfrac{{3,9}}{{6,25}} \times 100\%  = 62,4\% \\
\% {m_{{K_2}O}} = 37,6\% 
\end{array}\)