Calculate the volume of `1.00 mol L^(-1)` aqueous sodium hydroxide that is neutralized by `200 mL` of `2.00 mol L^(-1)` aqueous hydrochloric acid and the mass of sodium chloride produced. Neutralization reaction is,
`NaOH_((aq.))+HCl_((aq.))rarr NaCl_((aq.))+H_(2)O_((l))`

Explanation

Step I. Calculation of volume of NaOH solution neutralised
Volume of NaOH solution neutralised can be calculated by applying molarity equation
`M_(1)V_(1)=M_(2)V_(2)`
`(1 M)xxV_(1) = (2M)xx(200 mL)=400 mL`
Step II. Calculation of mass of `NaCl` produced
The neutralised reaction is :
`NaOH(aq)+underset(underset(36.5g)(("1 mol")))(HCl(aq))rarrunderset(underset(58.5g)(("1 mol")))(NaCl(aq))+H_(2)O(l)`
`:.` Mass of HCl in 200 ml of 2M solution can be calculated as follows :
molarity of solution `= ("Mass of HCl/Molar mass")/("Volume of solution in litres")`
`(2.0"mol L"^(-1))=("Mass of HCl")/(("36.5 g mol"^(-1))xx(0.2L))`
Mass of HCl `= (2.0 "mol L"^(-1))xx("36.5 g mol"^(-1))xx(0.2 L)=14.6 g`
Now, `36.5 g` of HCl produce NaCl after neutralisation = 58.5 g
`:.` 14.6 of HCl produce NaCl after neutralisation `= ((58.5g))/((36.5g))xx(14.6g)=23.4g`.