Đáp án:
8 C
9 A
10 B
Giải thích các bước giải:
\(\begin{array}{l}
8)\\
{C_6}{H_{12}}{O_6} + 2AgN{O_3} + 2N{H_3} + {H_2}O \to HOC{H_2}{(CHOH)_4}COON{H_4} + 2Ag + 2N{H_4}N{O_3}\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{18}}{{180}} = 0,1\,mol\\
{n_{Ag}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,2\,mol\\
H = 85\% \Rightarrow {m_{Ag}} = 0,2 \times 108 \times 85\% = 18,36g\\
9)\\
{C_6}{H_{12}}{O_6} + 2AgN{O_3} + 2N{H_3} + {H_2}O \to HOC{H_2}{(CHOH)_4}COON{H_4} + 2Ag + 2N{H_4}N{O_3}\\
{n_{Ag}} = \dfrac{{108}}{{108}} = 0,1\,mol\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{C_M}{C_6}{H_{12}}{O_6} = \dfrac{{0,05}}{{0,2}} = 0,25M\\
10)\\
{C_6}{H_{12}}{O_6} + 2AgN{O_3} + 2N{H_3} + {H_2}O \to HOC{H_2}{(CHOH)_4}COON{H_4} + 2Ag + 2N{H_4}N{O_3}\\
{n_{{C_6}{H_{12}}{O_6}}} = \dfrac{{54}}{{180}} = 0,3\,mol\\
{n_{Ag}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,6\,mol\\
H = 75\% \Rightarrow {m_{Ag}} = 0,6 \times 108 \times 75\% = 48,6g
\end{array}\)
