How much Ba(OH)2 is needed to neutralize 3.2 moles of H2SO4 and 4.5 moles of HCl mixed together in 755 mL of water?

Moles of H+ ions: 3.2*.755*2+ 4.5*.755

= .755(6.4+4.5)

moles of Ba(OH)2= 1/2 (moles H+ ions).

I am still a little confused, how did you get the .755(6.4+4.5)?

Look at the line above. I figured the moles from the two acids and added.

why did you multiply by .755 when we were given moles not molarity?

To find out how much Ba(OH)2 is needed to neutralize the given amounts of H2SO4 and HCl, we need to determine the balanced chemical equation for the neutralization reaction between Ba(OH)2 and the acids.

The balanced chemical equation for the neutralization reaction between Ba(OH)2 and H2SO4 is:
Ba(OH)2 + H2SO4 -> BaSO4 + 2H2O

The balanced chemical equation for the neutralization reaction between Ba(OH)2 and HCl is:
2HCl + Ba(OH)2 -> BaCl2 + 2H2O

From these balanced equations, we can see that 1 mole of Ba(OH)2 reacts with 1 mole of H2SO4, and 2 moles of HCl reacts with 1 mole of Ba(OH)2.

To determine the amount of Ba(OH)2 needed to neutralize the given amounts of H2SO4 and HCl, we can use the stoichiometry of the balanced chemical equations.

For H2SO4:
Ratio of Ba(OH)2 to H2SO4 = 1:1
Amount of Ba(OH)2 needed to neutralize 3.2 moles of H2SO4 = 3.2 moles

For HCl:
Ratio of Ba(OH)2 to HCl = 1:2
Amount of Ba(OH)2 needed to neutralize 4.5 moles of HCl = (4.5 moles) / 2 = 2.25 moles

Therefore, the total amount of Ba(OH)2 needed to neutralize the given mixture of H2SO4 and HCl is:
3.2 moles (from H2SO4) + 2.25 moles (from HCl) = 5.45 moles of Ba(OH)2

So, 5.45 moles of Ba(OH)2 is needed to neutralize the given mixture of H2SO4 and HCl.