a) Theo đề bài ta có : nHCl = \(\dfrac{200.1,825}{100.36,5}=0,1\left(mol\right)\)
Ta có PTHH :
\(ROH+HCl->RCl+H2O\)
0,1mol.....0,1mol
=> \(M_{ROH}=\dfrac{5,6}{0,1}=56\left(\dfrac{g}{mol}\right)=>M_R=56-17=39\left(\dfrac{g}{mol}\right)\)
b) Theo đề bài ta có : \(\left\{{}\begin{matrix}nKOH=\dfrac{200.8,4}{100.56}=0,3\left(mol\right)\\nCuSO4=\dfrac{200.16}{100.160}=0,2\left(mol\right)\end{matrix}\right.\)
Ta có PTHH :
\(2KOH+C\text{uS}O4->K2SO4+Cu\left(OH\right)2\downarrow\)
0,3mol.........0,15mol............0,15mol....0,15mol
Theo PTHH : \(nKOH=\dfrac{0,3}{2}mol< nCuSO4=\dfrac{0,2}{1}mol=>nCuSO4\left(d\text{ư}\right)\) ( tính theo nKOH)
Ta có : \(\left\{{}\begin{matrix}C\%_{C\text{uS}O4\left(d\text{ư}\right)}=\dfrac{\left(0,2-0,15\right).160}{0,3.56+200-0,15.98}.100\%\approx3,96\%\\C\%_{K2SO4}=\dfrac{0,15.174}{0,3.56+200-0,15.98}.100\%\approx12,9\%\end{matrix}\right.\)
