Đáp án + Giải thích các bước giải:

 a) $m_{H_2} = 20\% . 50 = 10g$

$n_{H_2} = \dfrac{m}{M} = \dfrac{10}{2} = 5\text{ mol}$
$m_{CH_4} = 50 - 10 = 40g$

$n_{CH_4} = \dfrac{m}{M} = \dfrac{40}{4 + 12} = \dfrac{40}{16} = 2,5\text{ mol}$
PTHH:
$2H_2 + O_2 \overset{t^o}{\longrightarrow} 2H_2O$(1)

$CH_4 + 2O_2 \overset{t^o}{\longrightarrow} CO_2 + 2H_2O$(2)

Theo PTHH(1), ta có:

$n_{O_2} = \dfrac{1}{2} n_{H_2} = \dfrac{1}{2} . 5 = 2,5\text{ mol}$

Theo PTHH(2), ta có:
$n_{O_2} = 2n_{CH_4} = 2 . 2,5 = 5\text{ mol}$

$\Rightarrow n_{O_2} = 2,5 + 5 = 7,5\text{ mol}$

$V_{O_2} = n . 22,4 = 7,5 . 22,4 = 168l$
$V_{O_2} = 25\% V_{kk} \Rightarrow V_{kk} = V_{O_2} : 25\% = 168 : 25\% = 672l$

b) Theo PTHH(1), ta có:
$n_{H_2O} = n_{H_2} = 5\text{ mol}$

Theo PTHH(2), ta có:
$n_{H_2O} = 2n_{CH_4} = 2 . 2,5 = 5\text{ mol}$

$\Rightarrow n_{H_2O} = 5 + 5 = 10\text{ mol}$
$m_{H_2O} = n . M = 10 . (2 + 16) = 10 . 18 = 180g$

$\text{Đáp án+Giải thích các bước giải:}$

$\text{Câu A:}$

$\text{PTHH:}$

$\text{(1) $2H_{2}$+$O_{2}$→$2H_{2}O$}$

$\text{(2) $CH_{4}$+$2O_{2}$→$CO_{2}$+$2H_{2}O$}$

$\text{Ta có: $mH_{2}$=$\dfrac{20×50}{100}$= 10 g}$

$\text{⇒ $nH_{2}$=$\dfrac{10}{2}$= 5 mol}$

$\text{⇒ $mCH_{4}$=50-10= 40 g}$

$\text{⇒ $nCH_{4}$=$\dfrac{40}{16}$= 2,5 mol}$

$\text{Theo PT(1) ta có: $nO_{2}$=$\dfrac{1}{2}$×5= 2,5 mol}$

$\text{Theo PT(2) ta có: $nO_{2}$=2×2,5= 5 mol}$

$\text{⇒ $nO_{2}$=2,5+5= 7,5 mol}$

$\text{$VO_{2}$=7,5×22,4= 168 lít}$

$\text{VKK=$\dfrac{168}{25}$×100= 672 lít}$

$\text{Câu B:}$

$\text{Theo PT(1) ta có: $nH_{2}O$=$nH_{2}$= 5 mol}$

$\text{Theo PT(2) ta có: $nH_{2}O$=2×2,5= 5 mol}$

$\text{⇒ $nH_{2}O$=5+5= 10 mol}$

$\text{⇒ $mH_{2}O$=10×18= 180 g}$

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