Đáp án + Giải thích các bước giải:

$a)n_{Na_2CO_3}=\dfrac{150.1,3.15,22\%}{106}\approx 0,28(mol)$

$n_{HCl}=\dfrac{43,8.25\%}{36,5}=0,3(mol)$

$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$

Xét tỉ lệ: $\dfrac{n_{HCl}}{2}<n_{Na_2CO_3}\Rightarrow Na_2CO_3$ dư

Theo PT: $\begin{cases} n_{NaCl}=0,3(mol)\\ n_{Na_2CO_3(dư)}=0,28-0,3:2=0,13(mol)\\ n_{CO_2}=0,15(mol)\end{cases}$

$\Rightarrow \begin{cases}m_{NaCl}=0,3.58,5=17,55(g)\\ m_{Na_2CO_3(dư)}=0,13.106=13,78(g)\\ m_{CO_2}=0,15.44=6,6(g) \end{cases}$

$\Rightarrow m_{\text{dd sau}}=150.1,3+43,8-6,6=232,2(g)$

$\Rightarrow \begin{cases} C\%_{Na_2CO_3(dư)}=\dfrac{13,78}{232,2}.100\%=5,93\%\\ C\%_{NaCl}=\dfrac{17,55}{232,2}.100\%=7,56\% \end{cases}$

$V_{\text{dd sau}}=V_{dd_{Na_2CO_3}}=150(ml)=0,15(l)$

$\Rightarrow \begin{cases} C_{M_{Na_2CO_3(dư)}}=\dfrac{0,13}{0,15}=0,87M\\ C_{M_{NaCl}}=\dfrac{0,3}{0,15}=2M \end{cases}$

$b)n_{KOH}=\dfrac{15,68}{56}=0,28(mol)$

$T=\dfrac{0,28}{0,15}=1,87$

$\Rightarrow $ Tạo 2 muối $K_2CO_3:x(mol)$ và $KHCO_3:y(mol)$

$CO_2+2KOH\to K_2CO_3+H_2O$

$CO_2+KOH\to KHCO_3$

$\Rightarrow \begin{cases} x+y=0,15\\ 2x+y=0,28 \end{cases}\Rightarrow \begin{cases} x=0,13\\ y=0,02 \end{cases}$

$\Rightarrow m_{muối}=0,13.138+0,02.100=19,94(g)$