For ammonia water and NaOH solution, add 10mL 0. 1mol? L- 1 AlCl3 solution produces aluminum hydroxide precipitation, but ammonia water is a weak electrolyte and partially ionized, while sodium hydroxide is a strong electrolyte and completely ionized. The amount of solute ammonia monohydrate in ammonia water and NaOH solution with the same volume and pH is greater than that of sodium hydroxide.

When both of them are insufficient, the precipitation amount of aluminum hydroxide is determined by ammonia monohydrate and sodium hydroxide, and the amount of ammonia monohydrate is greater than that of sodium hydroxide, so the precipitation in A is more than that in B;

When ammonia water is excessive and sodium hydroxide is insufficient, the amount of aluminum hydroxide generated in A is =10ml×10-3l/ml× 0.1mol? L- 1=0.00 1mol, and the amount of aluminum hydroxide generated in B is less than 0.00 1mol, so the precipitation in A is more than that in B;

When ammonia water is excessive and sodium hydroxide is just right, the precipitation generated depends on aluminum chloride, and the amount of aluminum chloride is equal, so A and B generate as much precipitation;

When both ammonia water and sodium hydroxide are in excess, because aluminum hydroxide can be dissolved in sodium hydroxide but not ammonia water, the aluminum hydroxide generated in A is 0.00 1mol, and the aluminum hydroxide generated in B is dissolved in excess sodium hydroxide, so the precipitation in A is more than that in B. 。

From the above analysis,

A, A's precipitation is not necessarily more than B's, so A is wrong.

B, A may have more precipitation than B, so B is correct.

C and A may have more precipitation than B, or as much, so C is wrong.

D. When both ammonia water and sodium hydroxide are excessive, aluminum hydroxide in ammonia water does not dissolve, while aluminum hydroxide in sodium hydroxide dissolves, so the phenomenon may be different, so D is wrong.

So choose B.