1)

 Phản ứng xảy ra:

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

Ta có: 

\({n_{{C_2}{H_5}OH}} = \frac{{2,3}}{{46}} = 0,05{\text{ mol}} \to {{\text{n}}_{C{O_2}}} = 2{n_{{C_2}{H_5}OH}} = 0,1{\text{ mol; }}{{\text{n}}_{{H_2}O}} = 3{n_{{C_2}{H_5}OH}} = 0,15{\text{ mol}}\)

\( \to {m_{C{O_2}}} + {m_{{H_2}O}} = 0,1.44 + 0,15.18 = 7,1{\text{ gam}}\)

\({n_{{O_2}}} = 3{n_{{C_2}{H_5}OH}} = 0,15{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,15.22,4 = 3,36{\text{ lít}} \to {{\text{V}}_{kk}} = 3,36.5 = 16,8{\text{ lít}}\)

2)

Gọi số mol \(CaC{O_3};{K_2}S{O_3}\) lần lượt là x, y.

\(CaC{O_3} + 2HCl\xrightarrow{{}}CaC{l_2} + C{O_2} + {H_2}O\)

\({K_2}S{O_3} + 2HCl\xrightarrow{{}}2KCl + S{O_2} + {H_2}O\)

\( \to 100x + 158y = 70,3{\text{ gam}}\)

Ta có: 

\({n_{S{O_2}}} + {n_{C{O_2}}} = x + y = \frac{{11,2}}{{22,4}} = 0,5{\text{ mol}}\)

Giải được:  x=0,15; y=0,35

\( \to {m_{CaC{O_3}}} = 0,15.100 = 15{\text{ gam}} \to {\text{\% }}{{\text{m}}_{CaC{O_3}}} = \frac{{15}}{{70,3}} = 21,34\%  \to \% {m_{{K_2}S{O_3}}} = 78,66\% \)