# In the reaction Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3, how many moles of Al(OH)_3 can be made with 2.3 moles of NaOH and excess Al_2(SO_4)3?

Feb 9, 2016

$0.767 m o l$
Since $6 m o l N a O H$ produces $2 m o l A l {\left(O H\right)}_{3}$, ie 3 times less, it implies that $2 , 3$ moles of $N a O H$ will produce $\frac{2.3}{3} = 0.767$ moles of $A l {\left(O H\right)}_{3}$.