Question

To neutralize 1.65g LiOH, how much .150 M HCl would be needed?

Answer 1

Well how many moles of lithium hydroxide you got?

Explanation:

We address the equation....

`LiOH(aq) + HCl(aq) rarr LiCl(aq) + H_2O(l)`

`"Moles of lithium hydroxide"=(1.65*g)/(23.95*g*mol^-1)=0.0689*mol`.

And we need an equivalent quantity of `HCl(aq)`.

`"Volume of HCl"=(0.0689*mol)/(0.150*mol*L^-1)xx10^3*mL*L^-1=45.9*mL`...