Em tham khảo nha:

\(\begin{array}{l}
10)\\
OH - C{H_2} - {(CHOH)_4} - CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to HO - C{H_2} - {(CHOH)_4} - COON{H_4} + 2N{H_4}N{O_3} + 2Ag\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{27}}{{180}} = 0,15\,mol\\
{n_{Ag}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,3\,mol\\
{m_{Ag}} = 0,3 \times 108 = 32,4g\\
12)\\
a)\\
OH - C{H_2} - {(CHOH)_4} - CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to HO - C{H_2} - {(CHOH)_4} - COON{H_4} + 2N{H_4}N{O_3} + 2Ag\\
b)\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{18}}{{180}} = 0,1\,mol\\
{n_{Ag}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,2\,mol\\
{m_{Ag}} = 0,2 \times 108 = 21,6g\\
13)\\
{C_6}{H_{12}}{O_6} \to 2{C_2}{H_5}OH + 2C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{40}}{{100}} = 0,4\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{{n_{C{O_2}}}}}{2} = 0,2\,mol\\
H = 75\%  \Rightarrow {m_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,2 \times 180}}{{75\% }} = 48g
\end{array}\)