Question

Given the equation: Al(OH)3 (s) + 3HCI (aq) AlCl3 (aq) +3H20 (1) a) Balance the equation b) What mass of HCl (aq) is required to react completely with 400 mg of Al(OH)3? Give your answer in GRAMS (0.56 g)
can someone please explain to ne how they got 0.56 grams

Answer 1

Let's take a step by step approach to why the answer is 0.56 g

01.Now we know that for every mole of aluminum hydroxide Al(OH)₃ three (3) moles of HCl are required

02. We have 400 mg of Al(OH)₃ and it's molar mass is 78

So in our 400 mg we have

0.4/78 moles = 0.00526 moles

03 as seen from step 1 we need 3 times the HCl moles than aluminum hydroxide

So for 0.005 moles of it we need

0.00526 ×3 = 0.01538 moles of HCl

04. 1 mole of HCl = 36.46 gram

So 0.01538 moles =

0.01538 ×36.46

= 0.56 grams

I hope this helps. If you have any query or want more detailed explanation, feel free to ask in the comments section below.