Question

Hòa tan 100g dd BaCl2 20,8% vào 125g K2CO3 13,8%

a) Viết PTHH

b) Tính C% dd sau PƯ

Answer 1

a) \(n_{BaCl_2}=\dfrac{100.20,8\%}{208}=0,1\left(mol\right)\)

\(n_{K_2CO_3}=\dfrac{125.13,8\%}{138}=0,125\left(mol\right)\)

PTHH: BaCl₂ + K₂CO₃ --> BaCO₃ + 2KCl

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,125}{1}\) => BaCl₂ hết, K₂CO₃ dư

PTHH: BaCl₂ + K₂CO₃ --> BaCO₃ + 2KCl

              0,1---->0,1--------->0,1---->0,2

m_{dd sau pư} = 100 + 125 - 0,1.197 = 205,3 (g)

\(\left\{{}\begin{matrix}C\%_{KCl}=\dfrac{0,2.74,5}{205,3}.100\%=7,26\%\\C\%_{K_2CO_3\left(dư\right)}=\dfrac{\left(0,125-0,1\right).138}{205,3}.100\%=1,68\%\end{matrix}\right.\)