Bạn bổ sung 200 (g) dd ở bài 1 là dd gì nhé.
Bài 2:
\(n_{K_2CO_3}=\dfrac{200.15\%}{138}=\dfrac{5}{23}\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{300.9,8\%}{98}=0,3\left(mol\right)\)
PT: \(K_2CO_3+H_2SO_4\rightarrow K_2SO_4+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{\dfrac{5}{23}}{1}< \dfrac{0,3}{1}\) ta được H2SO4 dư.
Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{K_2SO_4}=n_{CO_2}=n_{K_2CO_3}=\dfrac{5}{23}\left(mol\right)\)
⇒ nH2SO4 dư = 19/230 (mol)
Ta có: m dd sau pư = 200 + 300 - 5/23.44 = 490,43 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{\dfrac{5}{23}.174}{490,43}.100\%\approx7,7\%\\C\%_{H_2SO_4}=\dfrac{\dfrac{19}{230}.98}{490,43}.100\%\approx1,65\%\end{matrix}\right.\)