MgO + 2HCl → MgCl2 + H2O
\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)
\(m_{HCl}=100\times19,5\%=19,5\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{19,5}{36,5}=\dfrac{39}{73}\left(mol\right)\)
a) Theo PT: \(n_{MgO}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{MgO}=\dfrac{73}{390}n_{HCl}\)
Vì: \(\dfrac{73}{390}< \dfrac{1}{2}\) ⇒ HCl dư
Theo PT: \(n_{HCl}pư=2n_{MgO}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=\dfrac{39}{73}-0,2=\dfrac{122}{365}\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=\dfrac{122}{365}\times36,5=12,2\left(g\right)\)
b) Dung dịch sau phản ứng gồm: HCl dư và MgCl2
\(\Sigma m_{dd}=4+100=104\left(g\right)\)
Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,2\times95=19\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{104}\times100\%=18,27\%\)
\(C\%_{HCl}dư=\dfrac{12,2}{104}\times100\%=11,73\%\)
PTHH. MgO + 2HCl → MgCl2 + H2O
Theo bài ra ta có:
+ nMgO bđ=4/40=0,1(mol)
+ mHCl=100×19,5%=19,5(g)
⇒nHCl=19,5/36,5=39/73(mol)
Theo pthh ta có: nMgO = 1/2 . nHCl
Theo bài: nMgO=73/390 . nHCl
Vì: 73/390<1/2 ⇒ HCl dư
Theo pthh và bài ta có:
nHClpư=2 . nMgO=2×0,1=0,2(mol)
⇒nHCldư=39/73−0,2=122/365(mol)
⇒mHCldư=122/365×36,5=12,2(g)b) Dung dịch sau phản ứng gồm: HCl dư và MgCl2
Σmdd=4+100=104(g)
Theo PT: nMgCl2=nMgO=0,1(mol)
⇒mMgCl2=0,2×95=19(g)
⇒C%MgCl2=19/104×100%=18,27%
C%HCldư=12,2/104×100%=11,73%
nMgO = \(\dfrac{4}{40}\)= 0.1mol
mHCl = \(\dfrac{19,5.100}{100}\) = 19,5 g =>nHCl = \(\dfrac{19,5}{36,5}\) \(\approx\)0,5 mol
MgO +2 HCl ------> MgCl2 + H2O
0,1(hết);0,5(dư)---> 0,1 mol
=>C%MgCl2 = \(\dfrac{0,1.95}{4+100}.100\%\) = 9,134%
