Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{{C_2}{H_4}}} = 60\% \\
\% {V_{C{H_4}}} = 40\% \\
b)\\
{m_{CaC{O_3}}} = 80g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{n_{C{O_2}}} = \dfrac{m}{M} = \dfrac{{35,2}}{{44}} = 0,8mol\\
{n_{hh}} = \dfrac{V}{{22,4}} = \dfrac{{11,2}}{{22,4}} = 0,5mol\\
hh:{C_2}{H_4}(a\,mol);C{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,5\\
2a + b = 0,8
\end{array} \right.\\
\Rightarrow a = 0,3;b = 0,2\\
{V_{{C_2}{H_4}}} = 0,3 \times 22,4 = 6,72l\\
\% {V_{{C_2}{H_4}}} = \dfrac{{6,72}}{{11,2}} \times 100\% = 60\% \\
\% {V_{C{H_4}}} = 100 - 60 = 40\% \\
b)\\
CaC{O_3} \xrightarrow{t^0} CaO + C{O_2}\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,8mol\\
{m_{CaC{O_3}}} = n \times M = 0,8 \times 100 = 80g
\end{array}\)