VHCl = 200 ml = 0,2 (l)
=> nHCl = 1.0,2 = 0,2 mol
NaOH + HCl -> NaCl + H2O
x------->x------>x
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
y------------>2y----->y
- ta có :\(\left\{{}\begin{matrix}40x+74y=7,7\\x+2y=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1mol\\y=0,05mol\end{matrix}\right.\)
a) mNaOH = 0,1 . 40 = 4g
mCa(OH)2 = 0,05.74 = 3,7 g
b)mNaCl = 0,1 . 58,5 = 5,85 g
mCaCl2 = 0,05 . 111 = 5,55 g
c) CM(NaCl và CaCl2) = \(\dfrac{0,1+0,05}{0,2}\) = 0,75 M
NaOH + HCl → NaCl + H2O (1)
Ca(OH)2 + 2HCl → CaCl2 + 2H2O (2)
a) \(n_{HCl}=0,2\times1=0,2\left(mol\right)\)
Gọi \(x,y\) lần lượt là số mol của NaOH và Ca(OH)2
Theo PT1: \(n_{HCl}=n_{NaOH}=x\left(mol\right)\)
Theo PT2: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=2y\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}40x+74y=7,7\\x+2y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
Vậy \(n_{NaOH}=0,1\left(mol\right)\) và \(n_{Ca\left(OH\right)_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,1\times40=4\left(g\right)\)
\(m_{Ca\left(OH\right)_2}=0,05\times74=3,7\left(g\right)\)
\(\Rightarrow\%m_{NaOH}=\dfrac{4}{7,7}\times100\%=51,95\%\)
\(\%m_{Ca\left(OH\right)_2}=\dfrac{3,7}{7,7}\times100\%=48,05\%\)
b) Theo PT1: \(n_{NaCl}=n_{NaOH}=0,1\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,1\times58,5=5,85\left(g\right)\)
Theo PT2: \(n_{CaCl_2}=n_{Ca\left(OH\right)_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{CaCl_2}=0,05\times111=5,55\left(g\right)\)
c) \(\Sigma n_{ct}=0,1+0,05=0,15\left(mol\right)\)
\(\Rightarrow C_{M_{dd}}saupứ=\dfrac{0,15}{0,2}=0,75\left(M\right)\)
