Question

(2 mL of each)

Sodium Bromide, 0.1 M

Sodium hydroxide, 0.1 M

Sodium sulfide, 2 M

Sodium sulfate, 0.1 M

Potassium Chlorate, 0.1 M

Potassium iodate, 0.1 M

Ammonium chloride, 0.2 M

Barium chloride, 0.2 M

Lead(II) nitrate, 0.1 M

1. You are using 9 different chemicals for this laborartory experiment. For each possible reaction between the 9 chemicals, predict the products and write a balnced, complete molecular equation.

2. Look at the products you predicted for the first question. Create a three column table of a) insoluble precipate formed, b) gas formed, and c) no reaction. Place the reactions in the appropriate column(s).

Answer 1

Step-1 :-

9 Chemicals are as follows:-

Sodium Bromide, 0.1 M

Sodium hydroxide, 0.1 M

Sodium sulfide, 2 M

Sodium sulfate, 0.1 M

Potassium Chlorate, 0.1 M

Potassium iodate, 0.1 M

Ammonium chloride, 0.2 M

Barium chloride, 0.2 M

Lead(II) nitrate, 0.1 M

Step-2:-

Understanding possible combination :- Total 38 Combination we can get.

Step-3:-

Differentiating between:-

a) insoluble precipate formed, b) gas formed, and c) no reaction.

In Appropriate Columns:

a) Insoluble Precipiate Formed

Pb(NO3)2 . Lead forms a lot of insoluble compounds :
1) Pb(NO3)2(aq) + 2NaBr(aq) → PbBr2(s) + 2NaNO3 (aq)
2) Pb(NO3)2(aq) + 2NaOH(aq) → Pb(OH)2(s) + 2NaNO3(aq)
3) Pb(NO3)2(aq) + Na2S (aq) → PbS(s) + 2NaNO3(aq)
4) Pb(NO3)2(aq) + Na2SO4(aq) → PbSO4(s) + 2NaNO3(aq)
5) Pb(NO3)2(aq) + KClO3(aq) → No reaction
6) Pb(NO3)2(aq) + KIO3(aq) → Pb(IO3)2(s) + 2KNO3(aq)
7) Pb(NO3)2(aq) + 2NH4Cl(aq) → PbCl2(s) + 2NH4NO3(aq)
8) Pb(NO3)2(aq) + BaCl2(aq) → PbCl2(s) + Ba(NO3)2 (aq)
A lot of barium salts are insoluble.
9) BaCl2(aq) + 2NaOH(aq) → Ba(OH)2(s) + 2NaCl(aq)
10) BaCl2(aq) + Na2S (aq) → BaS(s) + 2NaCl(aq)
11) BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
12) BaCl2(aq) + NaBr(aq) → No reaction
13) BaCl2(aq) + KClO3(aq) → No reaction
14) BaCl2(aq) + 2KIO3(aq) → Ba(IO3)2(s) + 2KNO3(aq)
15) BaCl2(aq) + NH4Cl(aq) → NO reaction

b) NH4Cl reacts wit a base to give ammonia gas It also reacts with Na2S:
16) NH4Cl(s) + NaOH (aq) → NaCl(aq) + NH3(g) + H2O(l)
17)2NH4Cl (aq) + Na2S(aq) → NaCl(aq) + 2NH3(g) + H2S(g)

c) There are no reactions with NaBr , Na2SO4 , KClO3, KIO3 and BaCl2 to keep the numbering up to date I list these no reactions: all reactants are (aq)
18) NH4Cl + NaBr → NO reaction
19) NH4Cl + Na2SO4 → No reaction
20) NH4Cl + KClO3 → No reaction
21) NH4Cl+ KIO3 → No reaction
22) NH4Cl + BaCl2 → No reaction

Now all you have left are a number of Na and K salts . None of these react together , so you have
23) KIO3 + KClO3 → No reaction
24) KIO3 + Na2SO4 → No reaction
25) KIO3 + Na2S → No reaction
26) KIO3 + NaOH → NO reaction
27) KIO3 NaOH → No reaction
28) KClO3 + Na2SO4 → No reaction
29) KClO3 + Na2S → No reaction
30) KClO3 + NaOH → No reaction
31) KClO3 + NaBr → No reaction
32) Na2SO4 + Na2S → No reaction
33) Na2SO4 NaOH → No reaction
34) Na2SO4 + NaBr → No reaction
35) Na2S + NaOH → NO reaction
36) Na2S +NaOH → NO reaction
37) Na2S + NaBr → NO reaction
38) NaOH + NaBr → NO reaction.