Question

I've seen some people having trouble on this question and now I am also stuck on this. I've tried my best to answer but I don't think its right. Can someone give me an honest answer (No links please or fake answers)
How many grams of Al(OH)3 are produced from 3.00 g of AlCl3 with excess of NaOH?

Answer 1

Answer:

approximately 1.772 grams

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

$3 \times \frac{132}{78} = 1.772 \: g \: al(oh)3$