a) Hiện tượng: sau khi phản ứng có chất kết tủa
K2CO3 + CaCl2 → CaCO3↓ + 2KCl
b) \(m_{K_2CO_3}=207\times20\%=41,4\left(g\right)\)
\(\Rightarrow n_{K_2CO_3}=\dfrac{41,4}{138}=0,3\left(mol\right)\)
\(m_{CaCl_2}=222\times10\%=22,2\left(g\right)\)
\(\Rightarrow n_{CaCl_2}=\dfrac{22,2}{111}=0,2\left(mol\right)\)
Theo PT: \(n_{K_2CO_3}=n_{CaCl_2}\)
Theo bài: \(n_{K_2CO_3}=\dfrac{3}{2}n_{CaCl_2}\)
Vì \(\dfrac{3}{2}>1\) ⇒ K2CO3 dư
Theo PT: \(n_{CaCO_3}=n_{CaCl_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{CaCO_3}=0,2\times100=20\left(g\right)\)
\(m_{dd}saupư=207+222-20=409\left(g\right)\)
Theo PT: \(n_{K_2CO_3}pư=n_{CaCl_2}=0,2\left(mol\right)\)
\(\Rightarrow n_{K_2CO_3}dư=0,3-0,2=0,1\left(mol\right)\)
\(\Rightarrow m_{K_2CO_3}dư=0,1\times138=13,8\left(g\right)\)
Theo PT: \(n_{KCl}=2n_{CaCl_2}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{KCl}=0,4\times74,5=29,8\left(g\right)\)
\(m_{ct}saupư=29,8+13,8=43,6\left(g\right)\)
\(\Rightarrow C\%_{dd}saupư=\dfrac{43,6}{409}\times100\%=10,66\%\)
a) cho dd K2CO3 vào dd CaCl2 thì ta thấy có kết tủa trắng
b)
nK2CO3 = \(\dfrac{207.20}{100.138}\)= 0,3 mol
nCaCl2 = \(\dfrac{222.10}{100.111}\) = 0,2 mol
K2CO3 + CaCl2 -> 2KCl + CaCO3 \(\downarrow\)
0,3(dư); 0,2(hết) ->0,4 mol ->0,2 mol
md2 = 207+222 - 0,2. 100 = 449 g
=>C%KCl = \(\dfrac{0,4.74,5}{449}.100\%\approx6,64\%\)
C%K2CO3(dư) = \(\dfrac{0,1.138}{449}.100\%\approx3,07\%\)
