Đáp án:

Giải thích các bước giải:

 `Na_2CO_3+2HCl→NaCl+CO_2↑+H_2O`

`n_{CO_2}=\frac{3,36}{22,4}=0,15\ mol`

Theo PT, `n_{HCl}=2n_{CO_2}=0,3\ mol`

`V_{HCl}=0,15\ lít`

b) `n_{Na_2CO_3}=n_{CO_2}=0,15\ mol`

`m_{Na_2CO_3}=0,15.106=15,9\ gam`

`%m_{Na_2CO_3}=\frac{15,9}{20}.100%= 79,5%`

`%m_{NaCl}=20,5%`

c) `n_{NaCl}=n_{CO_2}=0,15\ mol`

`m_{NaCl}=0,15.58,5=8,775\ gam`

`m_{muối}=8,775+4,1=12,875\ gam`

a.

$n_{CO_2}=\frac{3,36}{22,4}=0,15(mol)$ 

$Na_2CO_3+2HCl→NaCl+CO_2+H_2O$

   0,15            0,3              0,15         ←  0,15                        (mol)     

-$V_{dd...HCl}=\frac{0,3}{2}=0,15(l)$ 

b.

-%$m_{Na_2CO_3}=\frac{0,15.106}{20}.100$%$=79,5$%

-%$m_{NaCl}=100$%$-$%$m_{Na_2CO_3}=100$%$-79,5$%$=20,5$%

c.

$m_{NaCl}=0,15.58,5=8,775(g)$

$m_{NaCl}..ban..đầu=20-0,15.106=4,1(g)$

$⇒m_{muối}=8,775+4,1=12,875(g)$