Portage Learning Chemistry CHEM 103 Unit 2

1. SrBr2 + (NH4)2CO3 → SrCO3 + 2 NH4Br 2. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2

BALANCE THE FOLLOWING EQUATIONS: 1. SrBr2 + (NH4)2CO3 ⟶ SrCO3 + NH4Br 2. FeS2 + O2 ⟶ Fe2O3 + SO2

1. 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O 2. 4 P + 5 O2 → 2 P2O5 3. 2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2

Balance each of the following equations by placing coefficients in front of each substance. 1. C4H10 + O2 → CO2 + H2O 2. P + O2 → P2O5 3. Al + H2SO4 → Al2(SO4)3 + H2

1. C7H5NOBr C 7 x 12.01 = 84.07 H 5 x 1.008 = 5.04 N 1 x 14.01 = 14.01 O 1 x 16.00 = 16.00 Br 1 x 79.90 = 79.90 199.02 amu %C = 84.07/199.02 x 100 = 42.24% %H = 5.04/199.02 x 100 = 2.53% %N = 14.01/199.02 x 100 = 7.04% %O = 16.00/199.02 x 100 = 8.04% %Br = 79.90/199.02 x 100 = 40.15%

Show the calculation of the percent of each element present in the following compound.Report your answer to 2 places after the decimal. 1. C7H5NOBr

1. MW of Na2SO4 = 2 (22.99) + (32.07) + 4(16.00) = 142.05 %Na = 45.98 / 142.05 x 100 = 32.37% %S = 32.07 / 142.05 x 100 = 22.58% %O = 64.00 / 142.05 x 100 = 45.05% 2. MW of Ca(NO3)2 = (40.08) + 2 (14.01) + 6 (16.00) = 164.1 %Ca = 40.08 / 164.1 x 100 = 24.42% %N = 28.02 / 164.1 x 100 = 17.07% %O = 96.00 / 164.1 x 100 = 58.50%

Calculate the percent of each element present in the following compounds: 1. Na2SO4 Na = __________ S = __________ O = __________ 2. Ca(NO3)2 Ca = __________ N = __________ O = __________

1. H+NO3- + K+OH- → K+NO3- + H2O = Double Replacement, 2 ionic → molecular 2. C7H8 + 9 O2 → 7 CO2 + 4 H2O = Combustion, Hydrocarbon + O2 → CO2 + H2O 3. 2 Al + 3 ZnCl2 → 3 Zn + 2 AlCl3 = Single Replacement, Metal displaces metal ion

Classify each of the following reactions as either: Combination Decomposition Combustion Double Replacement Single Replacement 1. H+NO3- + K+OH- → K+NO3- + H2O 2. C7H8 + 9 O2 → 7 CO2 + 4 H2O 3. 2 Al + 3 ZnCl2 → 3 Zn + 2 AlCl3

#1 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 (60 grams FeS2 / 119.99) = 0.5000 moles FeS2 0.5000 moles FeS2 x (8 / 4) = 1.00 mole SO2 1.00 mole SO2 x (64.07) = 64.1 grams of SO2 #2 NH4NO2 → N2 + 2 H2O (25 grams N2 / 28.02) = 0.8922 moles N2 0.8922 moles N2 x (1 / 1) = 0.892 mol of NH4NO2

Complete the following calculations and show your work. Be sure to give answers with the correct units. 1. How many grams of SO2 would be formed from 60 grams of FeS2 in the following reaction? 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 2. How many moles of NH4NO2 would be required to form 25 grams of N2 in the following reaction? NH4NO2 → N2 + 2 H2O

1.) % Composition of a compound is: 52.56% Fe 52.56% Fe ÷ 55.85 = 0.941 (smallest number of the set) 2.83% H 2.83% H ÷ 1.008 = 2.808 44.91% O 44.91% O ÷ 16.00 = 2.807 0.941 is the smallest of this set of numbers, so it is divided into each of the set of numbers Fe = 0.941 ÷ 0.941 = 1 Fe H = 2.808 ÷ 0.941 = 3 H FeH3O3 O = 2.816 ÷ 0.941 = 3 O 2.) % Composition of a compound is: 71.98% C 71.98% C ÷ 12.01 = 5.993 6.71% H 6.71% H ÷ 1.008 = 6.657 21.31% O 21.31% O ÷ 16.00 = 1.332 (smallest of the set) 1.332 is the smallest of this set of numbers, so it is divided into each of the set of numbers C = 5.993 ÷ 1.332 = 4.5 H = 6.657 ÷ 1.332 = 5 O = 1.332 ÷ 1.332 = 1 Since C is 4.5, multiply all numbers by 2 to get C9H10O2

Determine the empirical formula for the following compound % compositions: 1.) 52.56% Fe, 2.83% H, 44.91% O 2.) 71.98% C, 6.71% H, 21.31% O

1.) Ca^+2, 2 Cl^-1 + Al^+3, 3 NO3^-1 → Will not occur since neither set of ions forms a precipitate 2.) Pb^+2, 2 NO3^-1 + 2 Na^+1, 2 Br^-1 → PbBr2 ↓ + 2 Na^+1, 2 NO3^-1

Determine which of the following double replacement reactions will occur using the solubility rules and complete the equations that do: 1.) Ca^+2, 2 Cl^-1 + Al^+3, 3 NO3^-1 → 2.) Pb^+2, 2 NO3^-1 + 2 Na^+1, 2 Br^-1 →

1. Cl2 + 2 F^-1 → Will not occur since F2 is more active than Cl2 2. Ba + Fe^+2 → Fe + Ba+2 Will occur since Ba is more active than Fe 3. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O Combustion 4. HCl + NaOH → NaCl + H2O Double Replacement 5. Al + Fe(NO3)3 → Fe + Al(NO3)3 Single Replacement 6. 2 S + 3 O2 → 2 SO3 Combination 7. H2SO4 → SO3 + H2O Decomposition

Determine which of the following single replacement reactions will occur using the activity series and complete the equations that do: (Click here Download herefor the Activity Series.) 1.) Cl2 + 2 F^-1 → 2.) Ba + Fe^+2 → Balance the following equations and label each reactions as either (1) Combination, (2) Decomposition, (3) Combustion, (4) Double Replacement or (5) Single Replacement. 3. C6H12O6 + O2 → CO2 + H20 4. HCl + NaOH → NaCl + H2O 5. Al + Fe(NO3)3 → Fe + Al(NO3)3 6. S + O2 → SO3 7. H2SO4 → SO3 + H2O

1. MW of NH4NO3 = 80.052 moles = 10.0 / 80.052 = 0.125 mole 2. MW of Li3PO4 = 115.793 grams = 0.0500 x 115.793 = 5.79 grams

Number your answers, and show your work. Be sure to give answers with the correct units. Calculate the number of moles in 10.0 grams in the following compound: 1. NH4NO3 ____________________ Calculate the number of grams of each in 0.0500 mol in the following compound: 2. Li3PO4 ____________________

1. Ca3N2 Ca 3 x 40.08 = 120.24 N 2 x 14.01 = 28.02 120.24 + 28.02 = 148.26 2. Li3PO4 Li 3 x 6.941 = 20.823 P 1 x 30.97 = 30.97 O 4 x 16.00 = 64.00 20.823 + 30.97 + 64.00 = 115.793 3. NH4NO3 N 2 x 14.01 = 28.02 H 4 x 1.008 = 4.032 O 3 x 16.00 = 48.00 28.02 + 4.032 + 48.00 = 80.052

Number your answers, and show your work. Be sure to give answers with the correct units. Calculate the molecular weight of the following compounds: 1. Ca3N2 ____________________ 2. Li3PO4 ____________________ 3. NH4NO3 ____________________

2 C6H6 + 15 O2 → 12 CO2 + 6 H2O 20.6 g / (6 x 12.01 + 6 x 1.008) = 20.6 / 78.108 = 0.2637 mole x 12/2 = 1.58 mole CO2 1.582 mole CO2 x (12.01 + 2 x 16.00) = 69.6 g CO2

Show the balanced equation and the calculation of the number of moles and grams of CO2 formed from 20.6 grams of C6H6. Show your answers to 3 significant figures. C6H6 + O2 → CO2 + H2O

Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O ClO2: Each O is -2 (total is -4), so Cl is +4 NaClO2: Na is metal in group I = +1, each O is -2 (total is -4), so Cl is +3 C: is uncombined so C = 0 CaCO3: Ca is metal in group II = +2, each O is -2 (total is -6), so C is +4 Since Cl (on left side) is +4 and Cl (on right side) is +3: Cl changes by 1 Since C (on left side) is 0 and C (on right side) is +4: C changes by 4 Multiply Cl compounds by 4 and C compounds by 1 and after balancing other atoms = 1 Ca(OH)2 + 4 NaOH + 4 ClO2 + 1 C → 4 NaClO2 + 1 CaCO3 + 3 H2O

Show the balancing of the following redox equation, including the determination of the oxidation number (charge) of ONLY the atoms which are changing. Ca(OH)2 + NaOH + ClO2 + C → NaClO2 + CaCO3 + H2O

Mn 27.60/54.94 = 0.50 <--Smallest S 24.17/32.07 = 0.75 O 48.23/16.00 = 3.01 Mn 0.50/0.50 = 1 x 2 = 2 S 0.75/0.50 = 1.5 = 1 (1/2) x 2 = 3 O 3.01/0.50 = 6.0 x 2 = 12 Mn2S3O12

Show the calculation of the empirical formula for each compound whose elemental composition is shown below. 27.60% Mn, 24.17% S, 48.23% O

1. 2Al + 3C + 9O = 233.99 2. 8C + 6H + N + 4O + Cl = 215.59

Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. Al2(CO3)3 2. C8H6NO4Cl

1. Grams = Moles x molecular weight = 1.05 x 132.15 = 138.8 grams 2. Grams = Moles x molecular weight = 1.18 x 179.17 = 211.4 grams

Show the calculation of the number of grams in the given amount of the following substances. Report your answer to 1 place after the decimal. 1. 1.05 moles of (NH4)2SO4 2. 1.18 moles of C9H9NO3

1. Moles = grams / molecular weight = 12.0 / 310.18 = 0.0387 mole 2. Moles = grams / molecular weight = 15.0 / 229.61 = 0.0653 mole

Show the calculation of the number of moles in the given amount of the following substances. Report your answerto 3 significant figures. 1. 12.0 grams of Ca3(PO4)2 2. 15.0 grams of C9H8NO4Cl

1. Mn(NO3)2 + NaBiO3 + HNO3 → HMnO4 + Bi(NO3)3 + NaNO3 + H2O Mn(NO3)2: each NO3 is -1 (total is -2), so Mn is +2 HMnO4: H = +1, each O is -2 (total is -8), so Mn is +7 NaBiO3: Na is metal in group I = +1, each O is -2 (total is -6), so Bi is +5 Bi(NO3)3: each NO3 is -1 (total is -3), so Bi is +3 Since Mn (on left side) is +2 and Mn (on right side) is +7: Mn changes by 5 Since Bi (on left side) is +5 and Bi (on right side) is +3: N changes by 2 Multiply Mn compounds by 2 and Bi compounds by 5 and after balancing other atoms = 2 Mn(NO3)2 + 5 NaBiO3 + 16 HNO3 → 2 HMnO4 + 5 Bi(NO3)3 + 5 NaNO3 + 7 H2O 2. NaCrO2 + NaClO + NaOH → Na2CrO4 + NaCl + H2O NaCrO2: Na is metal in group I = +1, each O is -2 (total is -4), so Cr is +3 Na2CrO4: Na is metal in group I = +1 (total is +2), each O is -2 (total is -8), so Cr is +6 NaClO: Na is metal in group I = +1, each O is -2, so Cl is +1 NaCl: Na is metal in group I = +1, so Cl is -1 Since Cr (on left side) is +3 and Cr (on right side) is +6: Cr changes by 3 Since Cl (on left side) is +1 and Cl (on right side) is -1: Cl changes by 2 Multiply Cr compounds by 2 and Cl compounds by 3 and after balancing other atoms = 2 NaCrO2 + 3 NaClO + 2 NaOH → 2 Na2CrO4 + 3 NaCl + H2O

Show the calculation of the oxidation number (charge) of ONLY the atoms which are changing in the following redox equations and then show how those charges are used to balance the following redox equations: 1. Mn(NO3)2 + NaBiO3 + HNO3 → HMnO4 + Bi(NO3)3 + NaNO3 + H2O 2. NaCrO2 + NaClO + NaOH → Na2CrO4 + NaCl + H2O

Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO4 Na2HAsO3: Na is metal in group I = +1 (total is +2), H = +1, each O is -2 (total is -6), so As is +3 H3AsO4: H is +1 (total is +3), each O is -2 (total is -8), so As is +5 KBrO3: K is metal in group I = +1, each O is -2 (total is -6), so Br is +5 KBr: K is metal in group I = +1, so Br is -1

Show the calculation of the oxidation number (charge) of ONLY the atoms which are changing in the following redox equations. Na2HAsO3 + KBrO3 + HCl → NaCl + KBr + H3AsO4