4 grams of each pure hydrochloric acid and pure caustic soda are together dissolved in water. What weight of sodium chloride is obtained? 

Explanation

The stoichiometric equation is given as
`HCl+NaOHrarr NaCl+H_(2)O`
`"1 mole of HCl "-="1 mole of NaOH"`
36.5 grams of `HCI= 40` grams of NaOH
NaOH is the limiting reagent. Therefore, the weight of
NaCl obtained is limited by NaOH.
1 mole of `NaOH = 1` mole of NaCl
40 grams of `NaOH = 58.5` grams of NaCl
4 grams of `NaOH = ?`
The weight of sodium chloride obtained `=58.5xx(4)/(40)="5.85 grams"`