48)
Phản ứng xảy ra:
\({({C_6}{H_{10}}{O_5})_n} + n{H_2}O\xrightarrow{{{H^ + }}}n{C_6}{H_{12}}{O_6}\)
\({C_6}{H_{12}}{O_6} + 2AgN{O_3} + 3N{H_3} + {H_2}O\)
\(\xrightarrow{{}}{(C{H_2}OH)_5}COON{H_4} + 2Ag + 2N{H_4}N{O_3}\)
Ta có:
\({m_{{C_6}{H_{10}}{O_5}}} = 200.75\% = 150{\text{ gam}}\)
\( \to {n_{{C_6}{H_{10}}{O_5}}} = \frac{{150}}{{162}} = \frac{{25}}{{27}}{\text{ mol = }}{{\text{n}}_{{C_6}{H_{12}}{O_6lt}}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{{25}}{{27}}.80\% = \frac{{20}}{{27}}{\text{ mol}}\)
\( \to {n_{Ag}} = 2{n_{{C_6}{H_{12}}{O_6}}} = \frac{{20}}{{27}}.2 = \frac{{40}}{{27}}{\text{ mol}}\)
\( \to m = {m_{Ag}} = \frac{{40}}{{27}}.108 = 160{\text{ gam}}\)
Chọn \(A\).
49)
\(X\) là amin no đơn chức, mạch hở nên \(X\) có dạng \(C_nH_{2n+3}N\)
Đốt cháy \(X\)
\({C_n}{H_{2n + 3}}N + (1,5n + 0,75){O_2}\xrightarrow{{{t^o}}}nC{O_2} + (n + 1,5){H_2}O\)
\({C_n}{H_{2n + 3}}N + HCl\xrightarrow{{}}{C_n}{H_{2n + 4}}NCl\)
Ta có:
\({n_{HCl}} = 0,1.2 = 0,2{\text{ mol = }}{{\text{n}}_X}\)
\( \to {n_{{N_2}}} = \frac{1}{2}{n_X} = 0,1{\text{ mol}}\)
\( \to V = {V_{{N_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)
Chọn \(A\).
50)
Phản ứng xảy ra:
\({C_{12}}{H_{22}}{O_{11}} + {H_2}O\xrightarrow{{{H^ + }}}{C_6}{H_{12}}{O_6} + {C_6}{H_{12}}{O_6}\)
\({C_6}{H_{12}}{O_6} + {H_2}\xrightarrow{{Ni,{t^o}}}{C_6}{H_{14}}{O_6}\)
Ta có:
\({n_{{C_{12}}{H_{22}}{O_{11}}}} = \frac{{17,1}}{{12.12 + 22 + 16.1}} = 0,05{\text{ mol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}} = 2{n_{{C_{12}}{H_{22}}{O_{11}}}} = 0,05.2 = 0,1{\text{ mol}}\)
Vì hiệu suất là 75%.
\( \to {n_{{C_6}{H_{12}}{O_6}}} = 0,1.75\% = 0,075{\text{ mol}}\)
Ta có:
\({n_{{H_2}}} = {n_{{C_6}{H_{12}}{O_6}}} = 0,075{\text{ mol}}\)
\( \to {V_{{H_2}}} = 0,075.22,4 = 1,68{\text{ lít}}\)
Không có đáp án để chọn nha. Chỉ chọn đáp án gần đúng là \(C\) thôi.
