10)
Phản ứng xảy ra:
\({({C_6}{H_{10}}{O_5})_n} + n{H_2}O\xrightarrow{{{H^ + }}}n{C_6}{H_{12}}{O_6}\)
Ta có:
\({n_{{C_6}{H_{10}}{O_5}}} = \frac{{32,4}}{{12.6 + 10 + 16.5}} = 0,2{\text{ mol = }}{{\text{n}}_{{C_6}{H_{12}}{O_6}}}\)
\( \to {n_{Ag}} = 2{n_{{C_6}{H_{12}}{O_6}}} = 0,2.2 = 0,4{\text{ mol}}\)
\( \to {m_{Ag}} = 0,4.108 = 34,2{\text{ gam}}\)
Chọn C.
11)
Ta có:
\({n_{Ag}} = \frac{{21,6}}{{108}} = 0,2{\text{ mol}}\)
\( \to {n_{{C_6}{H_{12}}{O_6}{\text{ lt}}}} = \frac{1}{2}{n_{Ag}} = 0,1{\text{ mol}}\)
Vì hiệu suất phản ứng là 75%.
\( \to {n_{{C_6}{H_{12}}{O_6}}} = \frac{{0,1}}{{75\% }} = \frac{2}{{15}}\)
\( \to {m_{{C_6}{H_{12}}{O_6}}} = \frac{2}{{15}}.(12.6 + 12 + 16.6) = 24{\text{ gam}}\)
Chọn B.
12)
Phản ứng xảy ra:
\(HOOC - C{H_2} - C{H_2} - CH(N{H_2}) - COOH + 2NaOH\)
\(\xrightarrow{{}}NaOOC - C{H_2} - C{H_2} - CH(N{H_2}) - COONa + 2{H_2}O\)
Ta có:
\({n_{glutamic}} = \frac{{14,7}}{{147}} = 0,1{\text{ mol}} \)
\({n_{NaOH}} =2{n_{glutamic}} = 0,2{\text{ mol}}=n_{H_2O}\)
BTKL:
\({m_{axit\;{\text{glutamic}}}} + {m_{NaOH}} = {m_{muối}} + {m_{{H_2}O}}\)
\( \to 14,7 + 0,2.40 = {m_{muoi}} + 0,2.18\)
\( \to {m_{muối}} = 19,1{\text{ gam}}\)
Chọn A.