Đáp án:
\({m_{andehit}} = 14,6{\text{ gam}}\)
\({{\text{V}}_{KOH}} = 50{\text{ ml}}\)
Giải thích các bước giải:
Gọi công thức của 2 andehit có dạng \({C_n}{H_{2n + 1}}CHO\) x mol.
Hidro hóa andehit
\({C_n}{H_{2n + 1}}CHO + {H_2}\xrightarrow{{{t^o}}}{C_n}{H_{2n + 1}}C{H_2}OH\)
\(2{C_n}{H_{2n + 1}}C{H_2}OH + 2Na\xrightarrow{{}}2{C_n}{H_{2n + 1}}C{H_2}ONa + {H_2}\)
Ta có: \({n_{ancol}} = {n_{andehit}} = x{\text{ mol;}}{{\text{n}}_{{H_2}}} = \frac{1}{2}{n_{ancol}} = 0,5x = \frac{{3,36}}{{22,4}} = 0,15 \to x = 0,3\)
Oxi hóa 2 andehit
\(2{C_n}{H_{2n + 1}}CHO + O\xrightarrow{{}}2{C_n}{H_{2n + 1}}COOH\)
\({C_n}{H_{2n + 1}}COOH + KOH\xrightarrow{{}}{C_n}{H_{2n + 1}}COOK + {H_2}O\)
Ta có:
\({n_{KOH}} = {n_{{C_n}{H_{2n + 1}}COOK}} = 0,3{\text{ mol}} \to {{\text{M}}_{{C_n}{H_{2n + 1}}COOK}} = \frac{{30,8}}{{0,3}} = 14n + 1 + 44 + 39 \to n = \frac{4}{3}\)
\( \to {m_{andehit}} = 0,3.(14n + 1 + 29) = 14,6{\text{ gam}}\)
\({m_{KOH}} = 0,3.56 = 16,8{\text{ gam}} \to {m_{dd{\text{ KOH}}}} = \frac{{16,8}}{{28\% }} = 60{\text{ gam}} \to {{\text{V}}_{KOH}} = \frac{{60}}{{1,2}} = 50{\text{ ml}}\)