Đáp án:
\(\begin{array}{l}
b)\\
\% {V_{C{H_4}}} = 50\% \\
\% {V_{{C_2}{H_4}}} = 50\% \\
c)\\
{m_{{H_2}O}} = 7,2g\\
d)\\
{m_{CaC{O_3}}} = 30g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
b)\\
{n_{hh}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
a + b = 0,2(1)\\
a + 2b = 0,3(2)\\
\text{Từ (1) và (2)}\Rightarrow a = 0,1;b = 0,1\\
{V_{C{H_4}}} = 0,1 \times 22,4 = 2,24l\\
{V_{{C_2}{H_4}}} = 0,1 \times 22,4 = 2,24l\\
\% {V_{C{H_4}}} = \dfrac{{2,24}}{{4,48}} \times 100\% = 50\% \\
\% {V_{{C_2}{H_4}}} = \dfrac{{2,24}}{{4,48}} \times 100\% = 50\% \\
c)\\
{n_{{H_2}O}} = 2{n_{C{H_4}}} + 2{n_{{C_2}{H_4}}} = 0,4mol\\
{m_{{H_2}O}} = 0,4 \times 18 = 7,2g\\
d)\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,3mol\\
{m_{CaC{O_3}}} = 0,3 \times 100 = 30g
\end{array}\)