Đáp án:

`↓` 

Giải thích các bước giải:

`a/` `PTHH:`

           `Na_2CO_3+2HCl→2NaCl+CO_2↑+H_2O`

`b/` `m_{dd_{HCl}}=D.V=1,1.300=330(g)`

`m_{HCl}=``\frac{C%.m_{dd}}{100%}=\frac{14,6%.330}{100%}=48,18(g)` 

`n_{HCl}=``\frac{m}{M}=\frac{48,18}{36,5}=1,32(mol)`

`PTHH`             `Na_2CO_3``+``2HCl``→``2NaCl``+``CO_2↑``+``H_2O`

                                                `1,32`            `1,32`           `0,66`                            `mol`

`→n_{CO_2}=``\frac{1}{2}n_{HCl}=\frac{1,32.1}{2}=0,66(mol)`

`→V_{CO_2(đktc)}=n.22,4=0,66.22,4=14,784(l)`

`c/``m_{CO_2}=n.M=0,66.44=29,04(g)`

`m_{sau}=m_{dd_{Na_2CO_3}}+m_{dd_{HCl}}-m_{CO_2}=150+330-29,04=450,96(g)`

Theo pt, ta có: `n_{NaCl}=n_{HCl}=1,32(mol)`

`→C%_{NaCl}=``\frac{m_{NaCl}.100%}{m_{sau}}=\frac{1,32.58,5.100%}{450,96}≈17,12%`

`#Ly`

$a)$ 

PTHH: $2HCl$ $+$ $Na_2CO_3$ `->` $2NaCl$ $+$ $CO_2↑$ $+$ $H_2O$

$b)$

$m_{ddHCl}$ $=$ $1,1.300=330$ $(g)$
$m_{HCl}$ $=$ $\dfrac{330.14,6}{100}$ $=$ $48,18$ $(g)$

$n_{HCl}$ $=$ $\dfrac{48,18}{36,5}$ $=$ $1,32$ $(mol)$

Theo PT, $n_{CO_2}$ $=$ $\dfrac{1}{2}n_{HCl}$ $=$ $0,66$ $(mol)$

$V_{CO_2}$ $=$ $0,66.22,4=14,784$ $(l)$

$c)$
Theo PT, $n_{NaCl}$ $=$ $n_{HCl}$ $=$ $1,32$ $(mol)$

$m_{NaCl}$ $=$ $1,32.58,5=77,22$ $(g)$

$m_{ddNaCl}$ $=$ $330+150-(0,66.44)=450,96$ $(g)$

`C%_(NaCl` $=$ $\dfrac{77,22}{450,96}$`.100%` $≈$ $17,12$`%`