150ml = 0,15l
\(n_{HCl}=2.0,15=0,3\left(mol\right)\)
Pt : \(Al\left(OH\right)_3+3HCl\rightarrow AlCl_3+3H_2O|\)
1 3 1 3
0,1 0,3 0,1
a) \(n_{Al\left(OH\right)3}=\dfrac{0,3.1}{3}=0,1\left(mol\right)\)
⇒ \(m_{Al\left(OH\right)3}=0,1.78=7,8\left(g\right)\)
b) \(n_{AlCl3}=\dfrac{0,3.1}{3}=0,1\left(mol\right)\)
⇒ \(m_{AlCl3}=0,1.133,5=13,35\left(g\right)\)
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