\(n_{Fe_2O_3}=0,2(mol)\\ Fe_2O_3+6HCl \to 2FeCl_3+3H_2O\\ n_{HCl}=1,2(mol)\\ V_{ddHCl}=\frac{250}{1,25}=200(ml)=0,2(l)\\ CM_{HCl}=\frac{1,2}{0,2}=6M$\)

a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{BaCO_3}=\dfrac{68,95}{197}=0,35\left(mol\right)\\n_{HCl}=0,25\cdot3,2=0,8\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,35}{1}< \dfrac{0,8}{2}\) \(\Rightarrow\) Axit còn dư 

\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,35\cdot2=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)

b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\) 

250ml=0,25l
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0.2.........0.4..........0,2............0,2   (mol)
a)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(m_{MgCl_2}=0,2.95=19\left(g\right)\)
b)
\(C_{M_{HCl}}=\dfrac{0,4}{0,25}=1,6\left(M\right)\)

a/ \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:      0,2       0,4            0,2

\(m_{MgCl_2}=0,2.95=19\left(g\right)\)

b/ \(C_{M_{ddHCl}}=\dfrac{0,4}{0,25}=1,6M\)

`n_[H_2 SO_4]=[9,8]/98=0,1(mol)`

`=>C_[M_[H_2 SO_4]]=[0,1]/[0,25]=0,4(M)`

\(a.Fe+2HCl\rightarrow FeCl_2+H_2\\ b.n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ TheoPT:n_{HCl}=2n_{Fe}=0,2\left(mol\right)\\ \Rightarrow CM_{HCl}=\dfrac{0,2}{0,2}=1M\)