Chemistry 20 – Lesson 23 Reactions in Solution Practice problems 1. Hydrochloric acid is added to a solution of hydroxide.

2 HCl (aq) + Ba(OH)2 (aq)  BaCl2 (aq) + 2 HOH (l) (non–ionic) + – 2+ – 2+ – 2 H (aq) + 2 Cl (aq) + Ba (aq) + 2 OH (aq)  Ba (aq) + 2 Cl (aq) + 2 HOH (l) (total ionic) + – H (aq) + OH (aq)  HOH (l) (net ionic)

2. Magnesium metal is added to an aqueous solution of hydrogen .

Mg (s) + 2 HBr (aq)  MgBr2 (aq) + H2 (g) (non–ionic) + – 2+ – Mg (s) + 2 H (aq) + 2 Br (aq)  Mg (aq) + 2 Br (aq) + H2 (g) (total ionic) + 2+ Mg (s) + 2 H (aq)  Mg (aq) + H2 (g) (net ionic)

3. Calcium metal reacts with water.

Ca (s) + 2 HOH (l)  H2 (g) + Ca(OH)2 (aq) (non–ionic) 2+ – Ca (s) + 2 HOH (l)  H2 (g) + Ca (aq) + 2 OH (aq) (total ionic) 2+ – Ca (s) + 2 HOH (l)  H2 (g) + Ca (aq) + 2 OH (aq) (net ionic)

4. Aqueous solutions of potassium and are mixed.

K2SO4 (aq) + BaCl2 (aq)  BaSO4 (s) + 2 KCl (aq) (non–ionic) + 2 2+ – + – 2 K (aq) + SO4 (aq) + Ba (aq) + 2 Cl (aq)  BaSO4 (s) + 2 K (aq) + 2 Cl (aq) (total ionic) 2 2+ SO4 (aq) + Ba (aq)  BaSO4 (s) (net ionic)

2+ 5. An aqueous solution of washing soda, Na2CO3, is added to remove Mg (aq) from water.

2+ + 2 + Mg (aq) + 2 Na (aq) + CO3 (aq)  2 Na (aq) + MgCO3 (s) (total ionic) 2+ 2 Mg (aq) + CO3 (aq)  MgCO3 (s) (net ionic)

Dr. Ron Licht 23 - 1 www.structuredindependentlearning.com Assignment /39 1. Potassium metal reacts with water.

/3 K (s) + 2 HOH (l)  H2 (g) + 2 KOH (aq) (non–ionic) + – K (s) + 2 HOH (l)  H2 (g) + 2 K (aq) + 2 OH (aq) (total ionic) + – K (s) + 2 HOH (l)  H2 (g) + 2 K (aq) + 2 OH (aq) (net ionic)

2. A lead (II) acetate solution reacts with a sodium sulfide solution to yield a precipitate.

/3 Pb(CH3COO)2 (aq) + Na2S (aq)  2 NaCH3COO (aq) + PbS (s) (non–ionic) 2+  + 2–  + Pb (aq) + 2 CH3COO (aq) + 2 Na (aq) + S (aq)  2 CH3COO (aq) + 2 Na (aq) + PbS(s) (total) 2+ 2– Pb (aq) + S (aq)  PbS(s) (net ionic)

3. Solutions of sodium sulfate and barium bromide are added together.

/3 Na2SO4 (aq) + BaBr2 (aq)  BaSO4 (s) + 2 NaBr (aq) (non–ionic) + 2 2+ – + – 2 Na (aq) + SO4 (aq) + Ba (aq) + 2 Br (aq)  BaSO4 (s) + 2 Na (aq) + 2 Br (aq) (total ionic) 2 2+ SO4 (aq) + Ba (aq)  BaSO4 (s) (net ionic)

4. An aqueous solution of sodium carbonate is used to remove calcium from water.

2+ + 2 + /3 Ca (aq) + 2 Na (aq) + CO3 (aq)  2 Na (aq) + CaCO3 (s) (total ionic) 2+ 2 Ca (aq) + CO3 (aq)  CaCO3 (s) (net ionic)

5. A precipitate forms when potassium iodide is mixed with lead (II) nitrate.

/3 Pb(NO3)2 (aq) + 2 KI (aq)  2 KNO3 (aq) + PbI2 (s) (non–ionic) 2+  + – + – Pb (aq) + 2 NO3 (aq) + 2 K (aq) + 2 I (aq)  2 K (aq) + 2 NO3 (aq) + PbI2 (s) (total ionic) 2+ – Pb (aq) + 2 I (aq)  PbI2 (s) (net ionic)

6. A calcium nitrate solution is added to a solution of sodium carbonate.

/3 Ca(NO3)2 (aq) + Na2CO3 (aq)  2 NaNO3 (aq) + CaCO3 (s) (non–ionic) 2+  + 2– + – Ca (aq) + 2 NO3 (aq) + 2 Na (aq) + CO3 (aq)  2 Na (aq) + 2 NO3 (aq) + CaCO3 (s) (total) 2+ 2– Ca (aq) + CO3 (aq)  CaCO3 (s) (net ionic)

Dr. Ron Licht 23 - 2 www.structuredindependentlearning.com 7. A precipitate forms when iron (III) nitrate reacts with sodium phosphate.

/3 Fe(NO3)3 (aq) + Na3PO4 (aq)  FePO4 (s) + 3 NaNO3 (aq) (non–ionic) 3+  + 3 +  Fe (aq) + 3 NO3 (aq) + 3 Na (aq) + PO4 (aq)  FePO4 (s) + 3 Na (aq) + 3 NO3 (aq) (total ionic) 3+ 3 Fe (aq) + PO4 (aq)  FePO4 (s) (net ionic)

2+ 2 8. Pb (aq) + S (aq)  PbS (s) c 0.100 mol c? Pb2 L S2 /6 v  0.0580L v 0.100L Pb2 S2

A. calculate moles B. mole ratio C. calculate concentration mol n2  0.100 (0.0580L) nn22 0.00580mol Pb L S Pb c   S2 n 0.00580mol 11 0.100L Pb2 n 2 0.00580mol c 0.0580 mol S  S2 L 11 n 0.00580mol S2

  9. Cl2 (aq) + 2 I (aq)  2 Cl (aq) + I2 (s) mol c  0.120 L m? I I2 /6 v 2.50L I

A. calculate moles B. mole ratio C. calculate mass mol g n  0.120L (2.50L) nI n  m 0.150mol(253.80mol ) I 2  I I2 n 0.300mol 12 I m 38.1g I2 n 0.300mol I2  12 n 0.150mol I2

3+ – 10. Fe (aq) + 3 OH (aq)  Fe(OH)3 (s) c? c 0.0200 mol Fe3 OH L /6 v  0.800L v 0.00480L Fe3 OH

A. calculate moles B. mole ratio C. calculate concentration mol n  0.0200 (0.00480L) nn3 0.0000320mol OH L Fe OH c   Fe3 n 0.0000960mol 13 0.800L OH n 5 Fe3 0.0000960mol c 4.00 10 mol  Fe3 L 13 n 0.0000320mol Fe3

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