Đáp án:
a, $\text{$V_{CO_{2}}$ = 13,44 (l)}$
b, $\text{$C_{M_{HCl}}$ = 2,4 (M)}$
c, $\text{$m_{NaCl}$ = 70,2 (g) }$
Giải thích các bước giải:
$\text{C% = $\dfrac{m_{ct}}{m_{dd}}$ . 100}$
$\text{$\Leftrightarrow$ 40 = $\dfrac{m}{159}$ . 100}$
$\text{$\Leftrightarrow$ $m_{Na_{2}CO_{3}}$ = 63,6 (g)}$
$\text{$\Rightarrow$ $n_{Na_{2}CO_{3}}$ = $\dfrac{m}{n}$ = $\dfrac{63,6}{23.2 + 12 + 16.3}$ = $\dfrac{63,6}{106}$ = 0,6 (mol)}$
$\text{PT: $Na_{2}CO_{3}$ + 2HCl $\rightarrow$ 2NaCl + $CO_{2}$$\uparrow$ + $H_{2}O$}$
$\text{(Mol) 0,6 $\longrightarrow$ 1,2 $\longrightarrow$ 1,2 $\longrightarrow$ 0,6 $\longrightarrow$ 0,6}$
$\text{a, Theo PT ta có: $n_{CO_{2}}$ = $n_{Na_{2}CO_{3}}$ = 0,6 (mol)}$
$\text{$\Rightarrow$ $V_{CO_{2}}$ = $n_{CO_{2}}$ . 22,4 = 0,6 . 22,4 = 13,44 (l)}$
$\text{b, Theo PT ta có: $n_{HCl}$ = $2n_{Na_{2}CO_{3}}$ = 2 . 0,6 = 1,2 (mol)}$
$\text{$V_{dd_{HCl}}$ = 500 (ml) = 0,5 (l)}$
$\text{$\Rightarrow$ $C_{M_{HCl}}$ = $\dfrac{n_{HCl}}{V_{HCl}}$ = $\dfrac{1,2}{0,5}$ = 2,4 (M)}$
$\text{c, Theo PT ta có: $n_{NaCl}$ = $2n_{Na_{2}CO_{3}}$ = 2 . 0,6 = 1,2 (mol)}$
$\text{$\Rightarrow$ $m_{NaCl}$ = $n_{NaCl}$ . $M_{NaCl}$ = 1,2 . (23 + 35,5) = 1,2 . 58,5 = 70,2 (g) }$
CHÚC BẠN HỌC TỐT:333
