Calculate the volume of 1.0 M aqueos solution of sodium hydroxide that is neutralised by 200 mL of 2.0 M aqueous HCl acid and also calculate the mass of sodium chloride produced.

Molarity is moles per litre of solute.
Thus
No. of moles of Sodium hydroxide present in 1 L solution = 1 mole
No. of moles HCl present in 1 L solution = 2 moles

No. of moles of HCl present in 200 ml = (2/1000) x 200 = 0.400 moles

1 mole of NaOH neutralised by 1 mole of HCl
Therefore, to neutralise 0.4 moles of HCl ,NaOH required = 0.4 moles

1 mole of NaOH present in = 1 litre solution
0.4 moles are present in = 400 ml of NaOH
400 ml of NaOH is required to neutralise 200 ml of HCl

Since 0.4 moles of sodium chloride is formed
Therefore mass of 0.4 moles of NaCl = No. of moles x Molar mass = 0.4 mol x 40 g/mol = 16 g