`n_{KHCO_3}=\frac{20.10%}{100}=0,02(mol)`

`n_{HCl}=\frac{50.14,6%}{36,5}=0,2(mol)`

$\rm KHCO_3 +HCl \xrightarrow{} KCl +CO_2 +H_2O$ 

tỉ lệ 

$\dfrac{0,02}{1} <\dfrac{0,2}{1}$`->KHCO_3` hết 

$m_{dd-s pứ}=20+50-0,02.44=69,12(g)$ 

`C%_{KCl }=\frac{m_{KCl}}{m_{dd}}=\frac{74,5.0,02}{69,12}.100%≈2,16%`

`C%_{HCl(dư)}=\frac{(0,2-0,02).36,5}{69,12}.100%≈9,5%`

Đáp án:

$\rm n_{KHCO_3}=\dfrac{20\cdot 10\%}{100}=0,02\ (mol)\\n_{HCl}=\dfrac{50\cdot 14,6\%}{36,5}=0,2\ (mol)$

Phản ứng xảy ra:

$\rm KHCO_3+HCl\to KCl+CO_2+H_2O$

Nhận xét: $\rm n_{KHCO_3}<n_{HCl}\to HCl$ dư, mọi tính toán theo $\rm KHCO_3$.

$\rm n_{KCl}=n_{CO_2}=n_{KHCO_3}=0,02\ (mol)$

$\rm n_{HCl\ (du)}=0,2-0,02=0,18\ (mol)$

Nồng độ phần trăm dd sau phản ứng:

$\rm C\%_{KCl}=\dfrac{0,02\cdot 74,5}{20+50-0,02\cdot 44}\cdot 100\%=2,16\%\\C\%_{HCl\ (du)}=\dfrac{0,18\cdot 36,5}{20+50-0,02\cdot 44}\cdot 100\%=9,5\%$