$ n_{C_2H_2} = \dfrac{V}{22,4} = \dfrac{11,2}{22,4} = 0,5 (mol) \\ 2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O $$\\$ a, $ V_{O_2} = n . 22,4 = 0,5.\dfrac{5}{2} . 22,4 = 28 (l) \\ V_{kk} = \dfrac{28}{20}.100  = 140 (l) $$ \\$ b, $ m_{CO_2} = n . M = 0,5.\dfrac{4}{2} . 44 = 44 (g) $$\\$ c, $ CO_2 + Ca(OH)_2 \rightarrow CaCO_3↓ + H_2O \\ m_{CaCO_3} = n . M = 0,5 . \dfrac{4}{2} . 100 = 100 (g) $

Đáp án:

\(\begin{array}{l}
a)\\
{V_{{O_2}}} = 28l\\
{V_{kk}} = 140l\\
b)\\
{m_{C{O_2}}} = 44g\\
c)\\
{m_{CaC{O_3}}} = 100g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{n_{{C_2}{H_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
{n_{{O_2}}} = \dfrac{5}{2}{n_{{C_2}{H_2}}} = 1,25mol\\
{V_{{O_2}}} = 1,25 \times 22,4 = 28l\\
{V_{kk}} = 5{V_{{O_2}}} = 5 \times 28 = 140l\\
b)\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_2}}} = 1mol\\
{m_{C{O_2}}} = 1 \times 44 = 44g\\
c)\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 1mol\\
{m_{CaC{O_3}}} = 1 \times 100 = 100g
\end{array}\)