Đáp án + Giải thích các bước giải:

Câu 6:

$a\big)$

Đặt $V_{CH_4(đkc)}=x(l);V_{C_2H_4(đkc)}=y(l)$

$\to x+y=6,1975(1)$

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

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

Theo PT: $x+2y=V_{CO_2(đkc)}=9,916(2)$

Từ $(1)(2)\to x=2,479(l);y=3,7185(l)$

$\to V_{CH_4(đkc)}=2,479(l);V_{C_2H_4(đkc)}=3,7185(l)$

$b\big)$

Theo PT: $n_{O_2}=2n_{CH_4}+3n_{C_2H_4}$

$\to V_{O_2(đkc)}=2x+3y=16,1135(l)$

$c\big)$

$n_{C_2H_4}=\frac{3,7185}{24,79}=0,15(mol)$

$C_2H_4+Br_2\to C_2H_4Br_2$

Theo PT: $n_{Br_2}=n_{C_2H_4}=0,15(mol)$

$\to V_{dd\,Br_2}=\frac{0,15}{0,5}=0,3(l)$

Câu 7:

$a\big)$

Đặt $n_{CH_4}=x(mol);n_{C_2H_2}=y(mol)$

$\to x+y=\frac{12,395}{24,79}=0,5(1)$

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

$2C_2H_2+5O_2\xrightarrow{t^o}4CO_2+2H_2O$

Theo PT: $x+2y=n_{CO_2}=\frac{35,2}{44}=0,8(2)$

Từ $(1)(2)\to x=0,2(mol);y=0,3(mol)$

$\to V_{CH_4(đkc)}=0,2.24,79=4,958(l)$

$\to V_{C_2H_2(đkc)}=0,3.24,79=7,437(l)$

$b\big)$

Theo PT: $n_{O_2}=2x+2,5y=1,15(mol)$

$\to V_{O_2(đkc)}=1,15.24,79=28,5085(l)$

$c\big)$

$C_2H_2+2Br_2\to C_2H_2Br_4$

Theo PT: $n_{Br_2}=2y=0,6(mol)$

Đổi $600ml=0,6l$

$\to C_{M\,Br_2}=\frac{0,6}{0,6}=1M$

Câu 8:

$a\big)$

Đặt $n_{CH_4}=x(mol);n_{C_2H_4}=y(mol)$

$\to 16x+28y=15(1)$

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

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

Theo PT: $2x+2y=n_{H_2O}=\frac{21,6}{18}=1,2(2)$

Từ $(1)(2)\to x=0,15(mol);y=0,45(mol)$

$\to m_{CH_4}=0,15.16=2,4(g)$

$\to m_{C_2H_4}=0,45.28=12,6(g)$

$b\big)$

Theo PT: $n_{CO_2}=x+2y=1,05(mol)$

$\to V_{CO_2(đkc)}=1,05.24,79=26,0295(l)$

$c\big)$

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

Theo PT: $n_{CaCO_3}=n_{CO_2}=1,05(mol)$

$\to m_{CaCO_3}=1,05.100=105(g)$

Đáp án + Giải thích các bước giải:

 Câu 6:
a)  Gọi $x\text{ mol} = n_{CH_4}; y\text{ mol} = n_{C_2H_4}$

$n_{hh} = n_{CH_4} + n_{C_2H_4} = \dfrac{6,1975}{24,79} = 0,25\text{ mol}$
$\Rightarrow x + y = 0,25(1)$

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

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

$n_{CO_2} = \dfrac{9,916}{24,79} = 0,4\text{ mol}$

Theo PTHH:
$n_{CO_2} = n_{CH_4} + 2n_{C_2H_4} = 0,4\text{ mol}$
$\Rightarrow x + 2y = 0,4(2)$
Từ $(1)$ và $(2)$, ta có:
$\begin {cases} x + y = 0,25 \\ x + 2y = 0,4 \end {cases}$

$\Rightarrow \begin {cases} x = 0,1 \\ y = 0,15 \end {cases}$

$\Rightarrow V_{CH_4}=  0,1 . 24,79 = 2,479l$

$V_{C_2H_4}= 6,1975 - 2,479 = 3,7185l$

b) Theo PTHH:
$n_{O_2} = 2n_{CH_4} + 3n_{C_2H_4}=  2 . 0,1 + 3 . 0,15 = 0,65\text{ mol}$
$\Rightarrow V_{O_2}=  0,65 . 24,79 = 16,1135l$

c) PTHH:
$C_2H_4 + Br_2 \longrightarrow C_2H_4Br_2$

Theo PTHH:
$n_{Br_2} = n_{C_2H_4} = 0,15\text{ mol}$
$\Rightarrow V_{Br_2} = \dfrac{n}{C_M} =\dfrac{0,15}{0,5} = 0,3l$

Câu 7:

a) Gọi $x\text{ mol} = n_{CH_4}; y\text{ mol} = n_{C_2H_2}$

$n_{hh} = n_{CH_4} + n_{C_2H_2} = \dfrac{12,395}{24,79} = 0,5\text{ mol}$
$\Rightarrow x + y = 0,5(1)$
PTHH:
$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

$2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O$

$n_{CO_2} = \dfrac{35,2}{44} = 0,8\text{ mol}$

Theo PTHH:
$n_{CO_2} = n_{CH_4} + 2n_{C_2H_2} = 0,8\text{ mol}$

$\Rightarrow x + 2y = 0,8(2)$

Từ $(1)$ và $(2)$, ta có:
$\begin {cases} x + y = 0,5 \\ x + 2y = 0,8 \end {cases}$

$\Rightarrow \begin {cases} x = 0,2 \\ y = 0,3 \end {cases}$

$\Rightarrow V_{CH_4}=  0,2 . 24,79 = 4,958l$
$V_{C_2H_2} = 12,395 - 4,958 = 7,437l$

b) Theo PTHH:
$n_{O_2} = 2n_{CH_4} + \dfrac{5}{2}n_{C_2H_2} = 2 . 0,2 + \dfrac{5}{2} . 0,3 = 1,15\text{ mol}$

$\Rightarrow V_{O_2} = 1,15 .24,79 = 28,5085l$

c) Đổi: $500ml = 0,5l$
PTHH:
$C_2H_2 + 2Br_2 \longrightarrow C_2H_2Br_4$
Theo PTHH:
$n_{Br_2} = 2n_{C_2H_2}=  2 . 0,3 = 0,6\text{ mol}$
$\Rightarrow C_{MBr_2} = \dfrac{n}{V} = \dfrac{0,6}{0,6} = 1M$

Câu 8:
Gọi $x\text{ mol} = n_{CH_4}; y\text{ mol} = n_{C_2H_4}$

Ta có:
$m_{CH_4} + m_{C_2H_4} = 15$
$\Rightarrow 16x + 28y = 15(1)$
PTHH:
$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

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

$n_{H_2O} = \dfrac{21,6}{18} = 1,2\text{ mol}$

Theo PTHH:
$n_{H_2O} = 2n_{CH_4} + 2n_{C_2H_4} = 1,2\text{ mol}$

$\Rightarrow 2x + 2y = 1,2(2)$

Từ $(1)$ và $(2)$, ta có:
$\begin {cases} 16x + 28y = 15 \\ 2x + 2y = 1,2 \end {cases}$

$\Rightarrow \begin {cases} x = 0,15 \\ y = 0,45 \end {cases}$

$\Rightarrow m_{CH_4} = 0,15 . 16 = 2,4g$

$m_{C_2H_4} = 15 - 2,4 =12,6g$

b) Theo PTHH:
$n_{CO_2}= n_{CH_4}  + 2n_{C_2H_4} = 0,15 + 2 . 0,45 = 1,05\text{ mol}$

$\Rightarrow V_{CO_2} = 1,05 . 24,79 = 26,0295l$

c) PTHH:
$CO_2 + Ca(OH)_2 \longrightarrow CaCO_3 \downarrow + H_2O$

Theo PTHH:
$n_{CaCO_3} = n_{CO_2} = 1,05\text{ mol}$
$\Rightarrow m_{CaCO_3} = 1,05 . 100 = 105g$