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?

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

answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )