Chem 121 Module 2 Exam

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Jan 9, 2024

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QuesƟon 1 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 Answer: 1. Ca3N2 = 3 (40.08) + 2 (14.01) = 148.26 2. Li3PO4 = 3 (6.941) + (30.97) + 4 (16.00) = 115.793 3. NH4NO3 = 2 (14.01) + 4 (1.008) + 3 (16.00) = 80.052 QuesƟon 2 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 in 0.0500 mol in the following compound: 2. Li3PO4 Answer: 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 QuesƟon 3 Number your answers, and be sure to show your work. Calculate the % composiƟon of the following compounds: 1. Na2SO4
2. Ca(NO3)2 Answer: 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% QuesƟon 4 Number your answers, and be sure to show your work. Determine the empirical formula for the following compound % composiƟons: 1. 52.56% Fe, 2.83% H, 44.91% O 2. 71.98% C, 6.71% H, 21.31% O Answer: 1) % ComposiƟon 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 O = 2.807 ÷ 0.941 = 3 O FeH3O3
2.) % ComposiƟon 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, mulƟply all numbers by 2 to get: C9H10O2 QuesƟon 5 BALANCE THE FOLLOWING EQUATIONS: 1. SrBr2 + (NH4)2CO3 → SrCO3 + NH4Br 2. FeS2 + O2 → Fe2O3 + SO2 Answer: 1. SrBr2 + (NH4)2CO3 → SrCO3 + 2 NH4Br 2. 4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2 QuesƟon 6 Determine which of the following double replacement reacƟons will occur using the solubility rules and complete the equaƟons 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 → Your Answer: 1. Ca^+2, 2 Cl^-1 + Al^+3, 3 NO3^-1 → Will not occur since neither set of ions forms a precipitate
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2. Pb^+2, 2 NO3^-1 + 2 Na^+1, 2 Br^-1 → PbBr2 ↓ + 2 Na^+1, 2 NO3^-1 QuesƟon 7 Determine which of the following single replacement reacƟons will occur using the acƟvity series and complete the equaƟons that do: 1. Cl2 + 2 F-1 → 2. Ba + Fe+2 → Label each of the following reacƟons as either (1) CombinaƟon, (2) DecomposiƟon, (3) CombusƟon, (4) Double Replacement or (5) Single Replacement and then balance the equaƟon. 3. C6H12O6 + O2 → CO2 + H2O 4. HCl + NaOH → NaCl + H2O 5. Al + Fe(NO3)3 → Fe + Al(NO3)3 6. S + O2 → SO3 7. H2SO4 → SO3 + H2O Answer: 1. Cl2 + 2 F^-1 → Will not occur since F2 is more acƟve than Cl2 2. Ba + Fe^+2 → Fe + Ba+2 Will occur since Ba is more acƟve than Fe 3. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O CombusƟon 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 CombinaƟon 7. H2SO4 → SO3 + H2O DecomposiƟon QuesƟon 8 Show the calculaƟon of the oxidaƟon number (charge) of ONLY the atoms which are changing in the following redox equaƟons and then show how those charges are used to balance the following redox equaƟons:
1. Mn(NO3)2 + NaBiO3 + HNO3 → HMnO4 + Bi(NO3)3 + NaNO3 + H2O 2. KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O Answer: 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 leŌ side) is +2 and Mn (on right side) is +7: Mn changes by 5 Since Bi (on leŌ side) is +5 and Bi (on right side) is +3: N changes by 2 MulƟply Mn compounds by 2 and Bi compounds by 5 and aŌer balancing other atoms = 2 Mn(NO3)2 + 5 NaBiO3 + 16 HNO3 → 2 HMnO4 + 5 Bi(NO3)3 + 5 NaNO3 + 7 H2O 2.) KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O KMnO4: K group I metal = +1, each O is -2 (total = -8), so Mn = +7 MnSO4: each SO4 is -2, so Mn is +2 H2C2O4: each H = +1 (total = +2), each O is -2 (total = -8), so C total is +6, each C is +3 CO2: each O is -2 (total = -4), so C is +4 Since Mn (on leŌ side) is +7 and Mn (on right side) is +2: Mn changes by 5 Since each C (on leŌ side) is +3 and C (on right side) is +4: C changes by 1 x 2 (since H2C2O4) = 2 MulƟply Mn compounds by 2 and H2C2O4 compound by 5 and aŌer balancing other atoms (which requires placing a 10 in front of CO2) = 2 KMnO4 + 5 H2C2O4 + 3 H2SO4 → 2 MnSO4 + K2SO4 + 10 CO2 + 8 H2O QuesƟon 9
Complete the following calculaƟons 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 reacƟon? 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 reacƟon? NH4NO2 N2 + 2 H2O Answer: 1. 60 X 8 X 64.07 = 64.1 grams of SO2 119.99 4 2. 25 X 1 = 0.892 mol of NH4NO2 28.02 1 QuesƟon 10 Complete the following calculaƟons and show your work. Be sure to give answers with the correct units. 1. Show the calculaƟon of the mass percent of solute in a soluƟon made by dissolving 15.9 grams of Ca3(PO4)2 in 350 grams of water. 2. Show the calculaƟon of the molality of a soluƟon made by dissolving 15.9 grams of Ca3(PO4)2 in 400 grams of water. 3. Show the calculaƟon of the molarity of a soluƟon made by dissolving 15.9 grams of Ca3(PO4)2 to make 350 ml of soluƟon. 4. Show the calculaƟon of the mass of Ca3(PO4)2 needed to make 200 ml of a 0.128 M soluƟon. 5. Show the calculaƟon of the volume of 0.238 M soluƟon which can be prepared using 13.4 grams of Ca3(PO4)2. Answer: 1. Mass % = (g[solute] / g[solute]+ g[solvent]) x 100% Mass % = (15.9 / 15.9 + 350) x 100 = 4.35%
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2. molality = (g[solute] / MW) / (g[solvent]/ 1000) molality = (15.9 / 310.18) / (400 / 1000) = 0.128 m 3. Molarity = (g[solute] / MW) / (ml[solvent]/ 1000) Molarity = (15.9 / 310.18) / (350 / 1000) = 0.146 M 4. Molarity = (moles) / (ml[solvent]/ 1000) 0.128 = (moles) / (200 / 1000) Moles = 0.128 x 0.200 = 0.0256 Moles = (g[solute] / MW) 0.0256 = (g[solute]/ 310.18) g[solute]= 0.0256 x 310.18 = 7.94 g 5. moles[solute]= g[solute] / MW moles[solute] = 13.4 g / 310.18 = 0.0432 mol Molarity = moles / (mL /1000) 0.238 = 0.0432 / (mL / 1000) mL / 1000 = 0.0432 / 0.238 = 0.1815 mL = 0.1815 x 1000 = 182 mL