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Question

I've gotten some answers but I have no idea how I got them.  

A chemist prepares a chemical reaction between aluminum nitrate and sodium hydroxide. The chemist 

 uses 100.0 mL of each solution in the reaction. Each solution has a concentration of 0.250 M.

(a) Write the net ionic equation for the reaction that occurs in solution.

(b) What is the mass of the precipitate that will form? 

(c) On the provided beaker, draw a particulate diagram to represent the reaction that occurs. Consider the ratio of ions shown in solution. Spectator ions and any excess ions from the reaction should be shown in an appropriate ratio. 

(d) What volume, in mL, of the excess reactant would have been sufficient to produce the mass of precipitate calculated in part (b)? Assume that the solutions

Answer 1

a

molecular: Al(NO3)3(aq)+3NaOH(aq) → Al(OH)3(s)+3NaNO3(aq)

NET

Al⁺³(aq)+3OH^-1(aq)→ Al(OH)3(s)

b

^{100mL Al(NO3)3}/₁ * ^{0.250molAl(NO3)3}/_{1000mLAl(NO3)3} * ^1molAl(OH)3/_1molAl(NO3)3 * ^{78.01gAl(OH)3}/_1molAl(OH)3

1.95g Al(OH)3

^{100mL NaOH}/₁ * ^0.250molNaOH/_1000mLNaOH * ^1molAl(OH)3/_3molNaOH * ^{78.01gAl(OH)3}/_1molAl(OH)3

0.650 g Al(OH)3

smaller mass forms from limiting reactant, NaOH must be limiting.

c

Don't show any OH- ions in solution because it's limiting. In solution show Na+, Al3+, and NO3-, then draw the precipitate, Al(OH)3.

Show the 3 NO3- , 1 Al3+, 9 Na+, and 1 Al(OH)3

This is approximate. More calculations needed.

d

^{0.650 g Al(OH)3}/₁ * ^1molAl(OH)3/_{78.01gAl(OH)3} * ^1molAl(NO3)3/_1molAl(OH)3 * ^{1000mLAl(NO3)3}/_{0.250molAl(NO3)3}

33.3 mL Al(NO3)3