What volume of 0.128 M HCl is required to neutralize 2.94 g of Mg(OH)2?

Question

I'm using MV=MV but I keep getting this question wrong because of orders of magnitude. I got .394 if that's any help. If anyone can help me with the process that would be amazing. Thank you so much

Answer 1

you got .394 what?

what are those units?

basically what your trying to do is find out how many liters of HCl that is .128 molar you need to create MgCl + H20. its almost a simple equalization problem after that.

i think...

Answer 2

Mg(OH)2 + 2HCl --------> MgCl2 + 2H2O

First work out the number of moles of Mg(OH)2 you are starting with.

Moles = mass / molecular weight

molecular weight Mg(OH)2 = 24.31 + (2 x 16.00) + (2 x 1.008) = 58.326 g/mol

So moles Mg(OH)2 = 2.94 g / 58.326 g/mol

= 0.0504 moles of Mg(OH)2

Now, from the balanced equation you see that 2 moles of HCl react with 1 moles of Mg(OH)2. Therefore you need 2 x the number of moles of HCl as you have Mg(OH)2 to neutralise the solution.

moles HCl required = 2 x 0.0504 moles

= 0.1008 moles of HCl.

Now simply work out how many ml of 0.128 M HCl has this number of moles.

Molarity = moles / Litres

therefore litres = moles / Molarity

= 0.1008 moles / 0.128 M

= 0.7875 litres

= 788 mL

Wow, that's a lot. I've re-checked though... sure I've done it right