Đốt cháy 30ml rượu Etylic 92 độ

a) tính Vkk cần phản ứng (đkc). Biết D rượu = 0,8 (g/mol) và Vo2 = 20% Vkk

b) dẫn sản phẩm qua 300 gam dd NaOH 20%. Tính nồng độ % muối trong dd sau phản ứng

Đáp án:

 a)

\(V_{C_2^{}H_5^{}OH(Nguyen\,Chat)}^{} = 30.92\%  = 27,6\,ml\)

\(m_{C_2^{}H_5^{}OH}^{} = V.D = 27,6.0,8 = 22,08\,gam\)

\(n_{C_2^{}H_5^{}OH}^{} = \frac{{22,08}}{{46}} = 0,48\,mol\)

\(C_2^{}H_5^{}OH + 3O_2^{}\xrightarrow{{{t^o}}}2CO_2^{} + 3H_2^{}O\)

\(0,48\,\,\,\,\,\,\,\,\,\,\,\,1,44\)

\(V_{O_2^{}}^{} = 1,44.22,4 = 32,256\,lit\)

\(V_{O_2^{}}^{} = 20\% .V_{kk}^{}\)

\( \to V_{kk}^{} = \frac{{V_{O_2^{}}^{}}}{{20\% }} = \frac{{32,256}}{{20\% }} = 161,28\,lit\)

\(b)\)

\(m_{NaOH}^{} = 300.20\%  = 60\,gam\)

\(n_{NaOH}^{} = \frac{{60}}{{40}} = 1,5\,mol\)

\(n_{CO_2^{}}^{} = 2.n_{C_2^{}H_5^{}OH}^{} = 0,48.2 = 0,96\,mol\)

\(T = \frac{{n_{NaOH}^{}}}{{n_{CO_2^{}}^{}}} = \frac{{1,5}}{{0,96}} = 1,5625\)

\( \to Na_2^{}CO_3^{}\,va\,NaHCO_3^{}\)

\(Na_2^{}CO_3^{}:x\,mol\)

\(NaHCO_3^{}:y\,mol\)

\(2NaOH + CO_2^{} \to Na_2^{}CO_3^{} + H_2^{}O\)

\(\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\, \leftarrow \,x\)

\(NaOH + CO_2^{} \to NaHCO_3^{}\)

\(\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\, \leftarrow \,y\)

\(2{\text{x}} + y = 1,5\)

\(x + y = 0,96\)

\( \to x = 0,54\,mol;y = 0,42\,mol\)

\(C\% _{Na_2^{}CO_3^{}}^{} = \frac{{0,54.106}}{{300 + 0,96.44}}.100 = 16,72\% \)

\(C\% _{NaHCO_3^{}}^{} = \frac{{0,42.106}}{{300 + 0,96.44}}.100 = 10,3\% \)