what's the volume of 0.50 mol/L HCl needed to neutralize 0.075 L of 0.90 mol/L of Ba(OH)2.

Answer 1

The balanced equation for the reaction taking place is...

2HCl + Ba(OH)2 ==> BaCl2 + 2H2O

Find the moles of Ba(OH)2 present:

0.075 L x 0.90 mol/L = 0.0675 moles Ba(OH)2 present

Next, find moles HCl needed:

0.0675 mol Ba(OH)2 x 2 mol HCl/mol Ba(OH)2 = 0.135 mol HCl needed to react with the Ba(OH)2

Finally, find the volume of HCl needed:

(x L)(0.5 mol/L ) = 0.135 moles

x = 0.27 L = 270 mls HCl needed (to 2 significant figures)

Answer 2

Paige

I will model this for you using 0.60, 0.080, 0.95 as the numbers (instead of 0.50, 0.075, 0.90) then you can follow my model.

Moles of Ba (OH)₂ = 0.080 x 0.95

Equation HCl + Ba (OH)₂ = .....

So you need 0.080 x 0.95 moles of HCl to react

1 L contains 0.60 moles of HCl so you need 0.080 x 0.95 / 0.60 l = 0.127 l of HCl

Mike