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))`

Answer

B, C, D

Explanation

`NaOH(aq) + HCl(aq) to NaCl(aq) + H_(2)O(l)`
1000 mL of 2.0 mol `L^(-1)` contain NaOH = 2.0 mol
200 mL of 2.0 mol `L^(-1)` contain NaOH
`=2.0/1000 xx 200`
=0.4 mol
1 mol of NaOH is neutralised by 1 mol of HCl
0.4 mol of NaOH will be neutralised by 0.4 mol of HCl
1.0 mol of 1.0 mol `L^(-1)` is present in 1000 mL
0.4 mol of 1.0 mol `L^(-1)` present in `1000/1.0 xx 0.4`
= 400 mL
1 mol of NaOH on reaction with HCl give = 58.5 g of NaCl
0.4 mol of NaOH on reaction with HCl will give
`=58.5/1 xx 0.4`
= 23.4 g