Chem Chap. 4
3. Using the solubility principles given in Tab. 4.1, p. 115, predict which of these ionic compounds are soluble in water. A). MgCl2 B). CaCO3 C). Pb(OH)2
A
1. Balance each equation. A). __ Fe + __ CO2 __Fe2O3 + __CO B). __ Al2(SO4)3 + __ NaOH __ Na2SO4 + __ Al(OH)3
A). 2 Fe + 3CO2 Fe2O3 + 3 CO B). Al2(SO4)3 + 6NaOH 3Na2SO4 + 2Al(OH)3
5. In the reaction 2 Li(s) + F2(g) 2 LiF(s) A. Which species is oxidized and which is reduced? B. Which species is the oxidizing agent and which is the reducing agent?
A. Which species is oxidized and which is reduced? 2 Li 2 Li+ + 2e- 2 Li loss 2 e- , therefore, Li is oxidized F2 + 2e- 2 F- F2 gains 2 e-, therefore, F2 is reduced B. Which species is the oxidizing agent and which is the reducing agent? 2 Li are reducing agents F2 is an oxidizing agents.
4. In the reaction Pb(s) + 2 Ag+(aq) Pb2+(aq) + 2 Ag(S) A. Which species is oxidized and which is reduced? B. Which species is the oxidizing agent and which is the reducing agent?
A. Which species is oxidized and which is reduced? Pb Pb2+ + 2 e- Pb losses 2 e- , therefore, Pb is oxidized. 2 Ag+ + 2 e- 2 Ag 2 Ag+ gain 2 e-, therefore, 2 Ag+ is reduced B. Which species is the oxidizing agent and which is the reducing agent? Pb is a reducing agent 2 Ag+ are oxidizing agents.
8. Magnesium reacts with sulfuric acid. How many moles of H2 are produced by the complete reaction of 0.23 g of Mg with sulfuric acid. Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g)
0.23 g ? mole 1 mole Mg = 24.3 g conversion factor: 1 mole Mg 24.3 g Mg 0.23 g Mg x 1 mole Mg = 0.0095 mole Mg = 0.0095 mole H2 24.3 g Mg
7.1 Calculate the number of grams in 0.84 mole of 2-propanol, C3H8O
1 mole C3H8O = 3 x 12 g + 8 x 1 g + 16 g = 60 g conversion factors: 60 g C3H8O and 1 mole C3H8O 1 mole C3H8O 60 g C3H8O 0.84 mole C3H8O x 60 g C3H8O = 50.4 g C3H8O 1 mole C3H8O
7.2 Calculate the number of moles of Mg2+ ions in 8.32 mole 0f Mg3(PO4)2
1 mole Mg3(PO4)2 = 3 moles Mg2+ 8.32 mole Mg3(PO4)2 x 3 moles Mg2+ = 24.96 moles Mg2+ 1 mole Mg3(PO4)2
2. When a solution of ammonium chloride, NH4Cl, is added to a solution of lead(II) nitrate, Pb(NO3)2, a white precipitate, lead(II) chloride, PbCl2, forms. __NH4Cl(aq) + __Pb(NO3)2(aq) __PbCl2(s) + __NH4NO3(aq) Write a balanced net ionic equation for this reaction. Ammonium chloride, lead(II) nitrate and ammonium nitrate exist as dissociated ions in aqueous solution.
2 NH4Cl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2NH4NO3(aq) 2 NH4+(aq) + 2 Cl-(aq) + Pb2+(aq) + 2 NO3-(aq) PbCl2(s) + 2 NH4+(aq) + 2 NO3-(aq) Net Ionic equation: 2 Cl-(aq) + Pb2+(aq) PbCl2(s)
6. Calculate the formula weight (amu) of: A). KCl B). Na3(PO4)2
A). KCl = 39.1 amu + 35.5 amu = 74.6 amu B). Na3(PO4)2 = 3 x 23 amu + 2 x (31 amu + 4 x 16 amu)= 259 amu
9. Iron ore is converted to iron by heating it with coal (carbon) and oxygen according to the following equation: 2 Fe2O3(s) + 6 C(s) + 3 O2(g) 4 Fe(s) + 6 CO2(g) If the process is run until 3940 g of Fe is produced, how many grams of CO2 will also be produced?
Step 1: 1 mole Fe = 55.85 g conversion factor: 1 mole Fe 55.85 g Fe 3940 g Fe x 1 mole Fe = 70.55 moles Fe 55.85 g Fe Step 2: 4 moles Fe = 6 moles CO2 conversion factor: 6 mole CO2 4 moles Fe 70.55 moles Fe x 6 mole CO2 = 105.83 moles CO2 4 moles Fe Step 3: 1 mole CO2 = 12 g + 2 x 16 g = 44 g conversion factor: 44 g CO2 1 mole CO2 105.83 moles CO2 x 44 g CO2 = 4656.92 g CO2 1 mole CO2
10. Benzene (C6H6) reacts with bromine (Br2) to produce bromobenzene (C6H5Br) according to the following equation: C6H6(l) + Br2(g) C6H5Br(l) + HBr(g) If 60 g of benzene is mixed with 135 g of bromine, A). Which is the limiting reagent? B). How many g of bromobenzene are formed in the reaction?
A). Which is the limiting reagent? Step 1: 1 mole C6H6 = 6 x 12 g + 6 x 1 g = 78 g conversion factor: 1 mole C6H6 78 g C6H6 60 g C6H6 x 1 mole C6H6 = 0.769 mole C6H6 78 g C6H6 Step 2: 1 mole Br2 = 2 x 79.9 g = 159.8 g Br2 conversion factor: 1 mole Br2 159.8 g Br2 135 g Br2 x 1 mole Br2 = 0.845 mole Br2 159.8 g Br2 Therefore: C6H6 is the limiting reagent B). How many g of bromobenzene are formed in the reaction? C6H6(l) + Br2(g) C6H5Br(l) + HBr(g) 0.769 mole 0.769 mole (? g) 0.769 mole C6H6 0.769 mole C6H5Br 1 mole C6H5Br = 6 x 12+ 5 x 1g + 80 g =157 g conversion factor: 157g C6H5Br 1 mole C6H5Br 0.769 mole C6H5Br x 157 g C6H5Br = 120.7 g C6H5Br 1 mole C6H5Br