Answer to Question #24232 in Inorganic Chemistry for Ileana
2Al + 2NaOH + 6H2O = 2Na[Al(OH)4] + 3H2
Al is in excess, so only 0.37 mol of it will react.
And about H2. Amount if it is 0.37 * 3/2 = 0.555 moles
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