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

400 mL, 23.4 g

Explanation

`underset("(NaOH)")(M_(1)V_(1))=underset("(HCl)")(M_(2)V_(2))," i.e., "1.0xxV_(1)=2.0xx200orV_(1)=400mL`
`" "NaOH+HClrarrNaCl+H_(2)O`
`"200 mL of 2.0 M HCl "=(2.0)/(1000)xx200=0.4 mol`
`"1 mol of HCl produces NaCl = 1 mol"`
`therefore" 0.4 mol of HCl will produce NaCl = 0.4 mol "=0.4xx58.5g=23.4g.`