Đáp án `+` Giải thích các bước giải:
$$\rm a)$$$$\rm n_{HCl} = \dfrac{136,875.8\%}{36,5} = 0,3(mol)$$$$\rm PTHH: Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O$$Theo PT: $$\rm n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}n_{HCl} = 0,15(mol)$$$$\rm \Rightarrow m_{Na_2CO_3} = 0,15.106 = 15,9(g)$$$$\rm \Rightarrow m_{ddNa_2CO_3} = \dfrac{15,9}{20\%} = 79,5(g)$$$$\rm b)$$$$\rm m_{CO_2} = 0,15.44 = 6,6(g)$$$$\rm c)$$Theo PT: $$\rm n_{NaCl} = n_{HCl} = 0,3(mol)$$$$\rm m_{dd} = 79,5 + 136,875 - 6,6 = 209,775(g)$$$$\rm \Rightarrow C\%_{NaCl} = \dfrac{0,3.58,5}{209,775}.100\% = 8,366\%$$

Đáp án + Giải thích các bước giải:
$$\rm n_{HCl} =$$ `[mdd.C%]/[100%.M_[HCl]] =` $$\dfrac{136,875.8}{100.36,5}$$ $$\rm {= 0,3 (mol)}$$
$$\rm {PTHH: Na_2CO_3 + 2HCl → 2NaCl + H_2O + CO_2↑}$$
Theo PT:
$$\begin{cases} n_{Na_2CO_3} = n_{CO_2} = \dfrac{1}{2}.n_{HCl}=0,15(mol)\\n_{NaCl} = n_{HCl} = 0,3(mol)\\ \end{cases}$$
`a)`
$$\rm mdd_{Na_2CO_3} =$$ `[mct.100%]/[C%] =` $$\dfrac{0,15.106.100}{20}$$ $$\rm {= 79,5 (g)}$$
`b)`
$$\rm m_{CO_2} = 0,15.44 = 6,6 (g)$$
`c)`
$$\rm mdd_{sau p.ứ} = mdd_{Na_2CO_3} + mdd_{HCl} - m_{CO_2}$$
                     $$\rm {= 79,5 + 136,875 - 6,6}$$
                    $$\rm {= 209,775 (g)}$$
⇒ $$\rm C\%_{NaCl} =$$ `[mct.100%]/[mdd] =` $$\dfrac{0,3.58,5.100}{209,775}$$ `≈ 8,37%`