Đáp án + Giải thích các bước giải:

Đặt $n_{CH_4}=x(mol);n_{C_2H_4}=y(mol);n_{C_2H_2}=z(mol)$ trong $5,6l\,A$

$\to x+y+z=\frac{5,6}{22,4}=0,25(1)$

Bảo toàn $\pi:n_{Br_2}=y+2z=2.\frac{20}{160}=0,25$ (do $5,6l$ gấp đôi $2,8l$)

$\to x=z(2)$

Bảo toàn $C,H:n_{CO_2}=x+2y+2z(mol);n_{H_2O}=2x+2y+z(mol)$

$n_{NaOH(bđ)}=\frac{180.20\%}{40}=0,9(mol)$

$CO_2+2NaOH\to Na_2CO_3+H_2O$

Theo PT: $n_{NaOH(dư)}=0,9-2(x+2y+2z)(mol)$

Ta có $m_{dd\,spu}=m_{CO_2}+m_{H_2O}+m_{dd\,NaOH}$

$\to m_{dd\,spu}=44(x+2y+2z)+18(2x+2y+z)+180=80x+124y+106z+180(g)$

$\to C\%_{NaOH(dư)}=\frac{40(0,9-2x-4y-4z)}{80x+124y+106z+180}.100\%=2,75\%$

$\to 36-80x-160y-160z=2,2x+3,41y+2,915z+4,95$

$\to 82,2x+163,41y+162,915z=31,05(3)$

Từ $(1)(2)(3)\to x=z\approx 0,12(mol);y\approx 0,01(mol)$

$\%V_{CH_4}=\%V_{C_2H_2}=\frac{0,12}{0,12.2+0,01}.100\%=48\%$

$\%V_{C_2H_4}=100-48.2=4\%$

Đáp án:

$\%V_{CH_4}=\%V_{C_2H_2}=48\%$

$\%V_{C_2H_4}=4\%$

Giải thích các bước giải:

Trong $5,6l\ A:$

Đặt: $n_{CH_4}=x(mol),\ n_{C_2H_4}=y(mol),\ n_{C_2H_2}=z(mol)$

$\to n_A=x+y+z=\dfrac{5,6}{22,4}=0,25(mol)\ (1)$

Bảo toàn liên kết $\pi :$

$y+2z=2×\dfrac{2,8}{22,4}=0,25(mol)$

$x+y+z=y+2z$

$\to x=z\ (2)$

Bảo toàn nguyên tố $C:$

$n_{CO_2}=x+2y+2z(mol)$

Bảo toàn nguyên tố $H:$

$n_{H_2O}=2n_{CH_4}+2n_{C_2H_4}+n_{C_2H_2}(mol)$

$n_{NaOH\ \rm (bđ)}=\dfrac{180×20\%}{40}=0,9(mol)$

$2NaOH+CO_2\to Na_2CO_3+H_2O$

$n_{NaOH\ \rm (dư)}=0,9-2×(x+2y+2z)(mol)$

$m_{dd\ sau}=180+44(x+2y+2z)+18(2x+2y+z)(g)$

$C\%_{NaOH\ \rm (dư)}=\dfrac{40×(0,9-2×(x+2y+2z))}{180+44(x+2y+2z)+18(2x+2y+z)}×100\%=2,75\%$

$\to 82,2x+163,41y+162,915z=31,05(g)\ (3)$

Từ $(1)(2)(3)$ giải được:

$x=z=0,12(mol),\ y=0,01(mol)$

$\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,12}{0,25}=48\%$

$\%V_{C_2H_4}=100\%-48\%×2=4\%$