joestarjonathan128
Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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?

Sagot :

ankit1990

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

[tex]3 \times \frac{132}{78} = 1.772 \: g \: al(oh)3[/tex]

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.