100.0cm3 of 2.0mol dm-3 hydrochloric acid is mixed with 2.5g of magnesium oxide powder. The mixture is stirred until no further change occurs. The resulting solution still contains excess hydrochloric acid. Calculate the volume of 1.5mol dm -3 sodium hydroxide solution needed to neutralise the resulting solution.(Relative atomic mass:O=16; Mg=24)

Answer 1

MgO + 2HCl => H2O + MgCl2

mols MgO = grams/molar mass = ?
mols HCl = M x L = ?

The problem tells you that all of the MgO was used but not all of the HCl was used.
How much HCl was used. That's mols MgO x (2 mols HCl/1 mol MgO) = ? mols HCl.

How much HCl is left. Subtract initial mols HCl from mols HCl used. that's what's left. That will be neutralized with the NaOH as follows.

NaOH + HCl ==> H2O + NaCl
Moles NaOH needed = mols HCl you have in the final solution.
Then M = mols NaOH/L NaOH
You know M NaOH (M = mols/dm^3), you know mols, solve for L NaOH. Post your work if you get stuck.