Đáp án + Giải thích các bước giải:

$a\big)C_2H_4+Br_2\to C_2H_4Br_2$

$b\big)$

$V_{CH_4}=V_{khí\,thoát}=2,479(l)$

$\to V_{C_2H_4}=3,7185-2,479=1,2395(l)$

$c\big)$

$n_{C_2H_4}=\frac{1,2395}{24,79}=0,05(mol)$

Theo PT: $n_{Br_2}=n_{C_2H_4}=0,05(mol)$

$\to m_{dd\,Br_2}=\frac{0,05.160}{10\%}=80(g)$

$d\big)$

$n_{CH_4}=\frac{2,479}{24,79}=0,1(mol)$

$C_2H_4+3O_2\xrightarrow{t^o}2CO_2+3H_2O$

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

Theo PT: $\begin{cases} \sum n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,2(mol)\\ \sum n_{H_2O}=2n_{CH_4}+2n_{C_2H_4}=0,3(mol) \end{cases}$

$\to\begin{cases} V=V_{CO_2(đkc)}=0,2.24,79=4,958(l)\\ m=m_{H_2O}=0,3.18=5,4(g) \end{cases}$

$e\big)$

$CO_2+Ca(OH)_2\to CaCO_3\downarrow+H_2O$

Theo PT: $n_{CaCO_3}=n_{CO_2}=0,2(mol)$

$\to m_{CR}=m_{CaCO_3}=0,2.100=20(g)$