Question

How would I do this? A beaker contains 50.0 mL of 0.25 M aluminum nitrate solution. What is the minimum volume of 0.2 M sodium sulfide solution that must be added in order to precipitate out all of the aluminum ions from the solution?

Answer 1

The minimum volume of 0.2M sodium sulfide that will precipitate out aluminum from 50.0 mL, 0.25 M aluminum nitrate would be 0.094 L or 94 mL

Stoichiometric calculation

From the equation of the reaction:

$2Al(NO_3)_3 + 3Na_2S ---> Al_2S_3 + 6NaNO_3$

Mole ratio of Na2S and Al(NO3)3 = 3:2

Mole of 50.0 mL, 0.25 M Al(NO3)3 = 50/1000 x 0.25

                                                         = 0.0125 mole

Equivalent mole of Na2S = 3/2 x 0.0125

                                         = 0.0188 mole

Volume of 0.2M, 0.0188 mole Na2S = 0.0188/0.2

                                                          = 0.094 L or 94 mL

More on stoichiometric calculations can be found here: brainly.com/question/8062886