a) \(\left\{{}\begin{matrix}n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\\n_{HCl}=\dfrac{200.18,25\%}{36,5}=1\left(mol\right)\end{matrix}\right.\)
PTHH: MgO + 2HCl ---> MgCl2 + H2O
bđ 0,2 1
pư 0,2---->0,4
sau pư 0 0,6 0,2
b) \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,2.95}{200+8}.100\%=9,13\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,8.36,5}{200+8}.100\%=14,04\%\end{matrix}\right.\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\) \(n_{HCl}=\dfrac{200\times18,25}{100\times36,5}=1\left(mol\right)\)
PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,2 0,4 0,2 (mol)
TL: \(\dfrac{0,2}{1}\) < \(\dfrac{1}{2}\) \(\Rightarrow\) HCl dư
\(C\%_{HCl\left(dư\right)}=\dfrac{\left(1-0,4\right)\times36,5\times100}{8+200}\approx10,5\%\)
\(C\%_{MgCl_2}=\dfrac{0,2\times95\times100}{8+200}\approx9,13\%\)
