Đáp án+ Giải thích các bước giải:

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

$b)$

$n_{C_2H_2}$ = `{13,44}/{22,4} = 0,6(mol)`

Theo pt : $n_{CO_2}$ = $2n_{C_2H_2}$ = $1,2(mol)$

`->` $m_{CO_2}$ = $n.M = 1,2.44 = 52,8(g)$

Theo pt : $n_{H_2O}$ = $n_{C_2H_2}$ = $0,6(mol)$

`->` $m_{CO_2}$ = $n.M = 0,6.18 = 10,8(g)$

$c)$

Theo pt : $n_{O_2}$ = `5/2.`$n_{C_2H_2}$ = $1,5(mol)$

`->` $V_{KK}$ = $5.$$V_{O_2}$ = $5.1,5.22,4 = 168(l)$

$\text{Câu 3.}$

$\text{a)}$

$\text{2C$_2$H$_2$ + 5O$_2$ $\xrightarrow{\text{t°}}$ 4CO$_2$ + 2H$_2$O}$

$\text{b)}$

$\text{n$_{\text{C$_2$H$_2$}}$ = $\dfrac{\text{V}}{\text{22,4}}$ = $\dfrac{\text{13,44}}{\text{22,4}}$ = 0,6}$ ( $\text{mol}$ )

$\rightarrow$ $\text{n$_{\text{CO$_2$}}$ = $\dfrac{\text{0,6 . 4}}{2}$ = 1,2}$ ( $\text{mol}$ )

$\Rightarrow$ $\text{m$_{\text{CO$_2$}}$ = n . M = 1,2 . 44 = 52,8}$ ( $\text{g}$ )

$\rightarrow$ $\text{n$_{\text{H$_2$O}}$ = $\dfrac{\text{0,6 . 2}}{2}$ = 0,6}$ ( $\text{mol}$ )

$\Rightarrow$ $\text{m$_{\text{H$_2$O}}$ = n . M = 0,6 . 18 = 10,8}$ ( $\text{g}$ )

$\text{c)}$

$\text{n$_{\text{O$_2$}}$ = $\dfrac{\text{0,6 . 5}}{2}$ = 1,5}$ ( $\text{mol}$ )

$\text{V$_{\text{O$_2$}}$ = $\text{n . 22,4}$ =  $\text{1,5 . 22,4}$ = 33,6}$ ( $\text{l}$ )

$-$ Mà $\text{V$_{\text{O$_2$}}$ = $\dfrac{1}{5}$V$_{\text{kk}}$}$

$\Rightarrow$ $\text{V$_{\text{kk}}$ = V$_{\text{O$_2$}}$ . 5 = 33,6 . 5 = 168}$ ( $\text{l}$ )