Câu 2:

a)

${C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O$

b)

${n_{{C_2}{H_4}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol$

Theo PTHH: ${n_{{O_2}}} = 3{n_{{C_2}{H_4}}} = 0,9mol$

$ \Rightarrow {V_{{O_2}}} = 0,9.22,4 = 20,16(l)$

c)

$C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O$

$ \Rightarrow {n_{CaC{O_3}}} = {n_{C{O_2}}} = 2{n_{{C_2}{H_4}}} = 0,6mol$

$ \Rightarrow {m_{CaC{O_3}}} = 0,6.100 = 60g$

Câu 3:

a)

${C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}$

b)

${n_{B{r_2}}} = \dfrac{{32}}{{160}} = 0,2mol$

Theo PTHH: ${n_{{C_2}{H_4}}} = {n_{B{r_2}}} = 0,2mol$

$ \Rightarrow {V_{{C_2}{H_4}}} = 0,2.22,4 = 4,48(l)$

$ \Rightarrow {V_{C{H_4}}} = 6,72 - 4,48 = 2,24(l)$

c)

$\begin{gathered}   \% {V_{C{H_4}}} = \dfrac{{2,24}}{{6,72}}.100\%  = 33,33\%  \hfill \\    \Rightarrow \% {V_{{C_2}{H_4}}} = 100 - 33,33 = 66,67\%  \hfill \\ \end{gathered} $