Question

How many grams of Al(OH)3 are produced from 3.00 g of AlCl3 with excess of NaOH?

Answer 1

Answer:

1.772 gram is the approximate answer

Explanation:

molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole

the reaction is

AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl

from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3

implies 132g AlCl3 gives 78g Al(OH)3

Implies 3g AlCl3 gives

3*122/78 = 1.772 grams