A solution contains some or all of the ions Cu2 Cu2 , Al3 Al3 , K K , Ca2 Ca2 , Ba2 Ba2 , Pb2 Pb2 , and NH4 NH4 . The following tests were performed, in order, on the solution. Addition of 6 MM HClHCl produced no reaction. Addition of H2SH2S with 0.2 MM HClHCl produced a black solid. Addition of (NH4)2S(NH4)2S produced a white solid. Addition of (NH4)2HPO4(NH4)2HPO4 in NH3NH3 produced no reaction. The final supernatant when heated produced a purple flame. Identify which of these ions were present in the solution, which were absent, and for which ones no conclusions can be drawn. Drag the appropriate items to their respective bins.

Explanation:

The masses of the solids provide us with the following two equations.

g Al(C H NO) g Mg(C H NO) g 9 6 3 9 6 2 + = 7.815

g Al O g MgO g 2 3 + =1.002

With two equations and four unknowns, we need two additional equa-

tions to solve the problem. A conservation of mass requires that all the

aluminum in Al(C9H6NO)3 is found in Al2O3; thus

g Al O g Al(C H NO) mol Al

459.45 g Al(C 2 3 = 9 6 3

1

9 6

2 3

H NO) 2 3

g Al O

3 mol Al O

101 96

2

g Al O g Al(C H NO) 2 3 = 9 6 0 11096 3 .

Using the same approach, a conservation of mass for magnesium gives g MgO g Mg(C H NO) mol Mgg Mg(C H 9 6 29

= 1312.61 6 2 NO)g MgOmol Mg40.304

g MgO g Mg(C H NO) = 9 6 2 0.12893

Substituting the equations for g MgO and g Al2O3 into the equation for

the combined weights of MgO and Al2O3 leaves us with two equations

and two unknowns.

g Al(C H NO) g Mg(C H NO) g 9 6 3 9 6 2 + = 7.815

0 11096 0 12893 3 . . + g Al(C H NO) g Mg(C H NO) 9 6 9 6 2 =1.002 g

Multiplying the first equation by 0.11096 and subtracting the second

equation gives

− 0. . 01797 g Mg(C H NO) g = −0 1348 9 6 2

g Mg(C H NO) g 9 6 2 = 7.501

g Al(C H NO) g g Mg(C H NO) 9 6 3 9 6 2 = − 7. . 815 7 501 = 0.314 g

Now we can finish the problem using the approach from Example 8.1. A

conservation of mass requires that all the aluminum and magnesium in the

sample of Dow metal is found in the precipitates of Al(C9H6NO)3 and the

Mg(C9H6NO)2. For aluminum, we find that

0 314 1

3 . g Al(C H NO) mol Al

459.45 g Al(C H 9 696NO)g Almol Al g Al3

26 982 = 0 01844 . .

0 01844

100 3 02 . . % g Al

0.611 g sample

= w/w Al

and for magnesium we have

7 501 1 2 . g Mg(C H NO) mol Mg

312.61 g Mg(C H 9 696NO)g MgmolMgg Mg2

24 305 = 0 5832 . .

0.5832 g Mg

0.611 g sample

= 100 95. %5 w/w Mg

answer; i believe the correct answer is (solid gold);

indium, and thallium have three electrons i think