a) CaO + 2HCl -> CaCl2 + H2O
----0,18---0,36----0,18-----0,18----
b) nCaO = 2,4/56 ≃ 0,43 (mol)
nHCl = 0,06.6 = 0,36 (mol)
=> CaO dư
mH2O = 0,18.18 = 3,24 (g)
c) CMCaCl2 = n/V = 0,18/0,06 = 3 (M)
Vậy ...
Lâu chưa làm có sai sót mong bạn thông cảm.
a) CaO + 2HCl → CaCl2 + H2O
b) \(n_{CaO}=\dfrac{2,4}{56}=\dfrac{3}{70}\left(mol\right)\)
\(n_{HCl}=0,06\times6=0,36\left(mol\right)\)
Theo PT: \(n_{CaO}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{CaO}=\dfrac{5}{42}n_{HCl}\)
Vì \(\dfrac{5}{42}< \dfrac{1}{2}\) ⇒ HCl dư
Theo PT: \(n_{H_2O}=n_{CaO}=\dfrac{3}{70}\left(mol\right)\)
\(\Rightarrow m_{H_2O}=\dfrac{3}{70}\times18=0,77\left(g\right)\)
c) Dung dịch sau phản ứng gồm: HCl dư, CaCl2
Theo PT: \(n_{HCl}pư=2n_{CaO}=2\times\dfrac{3}{70}=\dfrac{3}{35}\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=0,36-\dfrac{3}{35}=\dfrac{48}{175}\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}dư=\dfrac{48}{175}\div0,06=4,57\left(M\right)\)
Theo PT: \(n_{CaCl_2}=n_{CaO}=\dfrac{3}{70}\left(mol\right)\)
\(\Rightarrow C_{M_{CaCl_2}}=\dfrac{3}{70}\div0,06=0,71\left(M\right)\)
