Acid + alkali

First of all, write down the chemical equation of the reaction between lead(II) nitrate solution and sodium hydroxide solution first.

Pb(NO3)2 (aq) + 2NaCl (aq) → PbCl2 (s) + 2NaNO3 (aq)

From the equation, the molar ratio of Pb(NO3)2 : NaCl = 1 : 2

So, in order to give the greatest amount of precipitate, the above molar ratio should be used.


For A, obviously, sodium chloride solution is a limiting agent since all of this chemical is being used up. And only 5 cm^3 of lead (II) nitrate solutino has been used. So, the volume of the precipitate formed is 5 cm^3.

For B, sodium chloride solution is the limiting factor. Only 5 cm^3 of precipitate is formed.

For C, the limiting agent is still sodium chloride solution. However, this time, 20 cm^3 of NaCl solution is used and 10 cm^3 of Pb(NO3)2 solution is reacted. So the resultant volume of the precipitate is 10 cm^3.

For D, the limiting agent is still sodium chloride. 15 cm^3 of sodium chloride is reacted with 7.5 cm^3 of Pb(NO3)2. So the precipitate has a volume of 7.5 cm^3.


Therefore, C is the answer.


2007-06-17 11:04:19 補充：
Because referring to the chemical equation, 2 parts of NaCl are needed while 1 part of Pb(NO3)2 is needed. Therefore, by calculation, 5 cm^3 of Pb(NO3)2 is needed while 10 cm^3 of NaCl is needed. So, All of the NaCl has been reacted while Pb(NO3)2 has not. So NaCl is a limiting agent.