3Ba(OH)\(_2\)+ Al\(_2\)(SO\(_4\))\(_3\)------> 3 BaSO\(_4\)+ 2Al(OH)\(_3\)
Ta có ;n\(_{Ba\left(OH\right)_2}\)= 0,15.0,1=0,015( mol)
n\(_{Al_2}\left(SO_4\right)_3\)= 0,1.0,1=0,01 (Mol)
=> Ba(OH)\(_2\)hết
Theo PTHH: n\(_{BaSO_4}\)=n\(_{Ba\left(OH\right)_2}\)=0,015 (mol)
m\(_{BaSO_4}\)= 0,015. 233=3,495 (g)
n\(_{Al\left(OH\right)_3}\)= \(\frac{2}{3}\)n\(_{Ba\left(OH\right)_2}\)=0,01 mol
m\(Al\left(OH\right)_3\)= 0,01.78=0,78 g
\(n_{Ba\left(OH\right)_2}=0,1.0,15=0,015\left(mol\right);n_{Al_2\left(SO_4\right)_3}=0,1.0,1=0,01\left(mol\right)\)
\(PTHH:3Ba\left(OH\right)_2+Al_2\left(SO_4\right)_3\rightarrow3BaSO_4\downarrow+2Al\left(OH\right)_3\downarrow\)
(mol) 3 1 3 2
(mol) 0,015 \(5.10^{-3}\) 0,015 0,01
\(TL:\frac{0,015}{3}< \frac{0,01}{1}\rightarrow Al_2\left(SO_4\right)_3.du\)
\(m_{kt}=0,015.233+0,01.79=4,285\left(g\right)\)