Answer to Question #30676 in Other Chemistry for Kel
6NaOH + 2Al ---> 2Na3AlO3 + 3H2
For answering this question you need to find amount of every reactant.
n=m/Mw, where Mw is molecular weight of compound
n of NaOH is 100/40 = 2.5 mol
n of Al is 100/27=3.7 mol
The ratio of amounts of NaOH : Al, must be 6:3 or 2: 1.
2.5 mol of NaOH can react with 2.5/2= 1.25 mol of Al, but you have 3.7,so NaOH is the limiting reagent.
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