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la. Convert 4.27 x 1027 atoms of He to moles G: 4.27 x 1027 atoms R: amount in moles A: Use n = M. where NA = 6.022 x 1023 S 4.27 x 1027 n = 6.022 x 1023 ~ 7091.19 mol S: The amount of helium is approximately 7091.19 mol. 1b. Convert 2.91 x 1025 formula units of Ba(OH)2 to moles G: 2.91 x 1025 formula units R: amount in moles A: n = NA S 2.91 x 1023 n= 6.022 x 1023 ~ 0.4833 mol S: The amount of Ba(OH)2 is approximately 0.483 mol. 2a. Convert 14.8 mol of BaCro4 to grams G: 14.8 mol, molar mass of BaCro4 = 253.3 g/mol R: mass in grams A: m = n X M S m = 14.8 x 253.3 = 3748.84 g S: The mass of BaCro4 is approximately 3748.84 g. 2b. Convert 8.43 x 1025 molecules of PbO2 to grams G: 8.43 x 1025 molecules, Molar mass of PbO2 = 239.2 g/mol R: mass in grams A 1. n = N 2. m =n x M S: 8.43 x 1023 n = 6.022 x 1023 ~ 1.4 mol m = 1.4 x 239.2 = 334.88 g S: The mass of PbO2 is approximately 334.88 g. 3a. Convert 1.23 mol of (NH4)2CO: to number of molecules G: 1.23 mol R: number of molecules A: N = n X NA S: N = 1.23 x 6.022 x 1023 = 7.41 x 1023 S: The number of molecules is approximately 7.41 x 1023.mass o 2b. Convert 8.43 x 1025 molecules of PbO2 to grams G: 8.43 x 1025 molecules, Molar mass of PbO2 = 239.2 g/mol R: mass in grams A: 1. n = N 2. m =n x M S. 8.43 x 1023 n = 6.022 x 1023 ~ 1.4 mol m = 1.4 x 239.2 = 334.88 g S: The mass of PbO2 is approximately 334.88 g. 3a. Convert 1.23 mol of (NH4)2CO: to number of molecules G: 1.23 mol R: number of molecules A: N = n X NA S: N = 1.23 x 6.022 x 1023 = 7.41 x 1023 S: The number of molecules is approximately 7.41 x 1023. 3b. Convert 15.6 g of MgCl2 to number of molecules G: 15.6 g, Molar mass of MgCl2 = 95.2 g/mol R: number of molecules A: 1. n = 2. N =n XNA S; 15.6 n : ~0.164 mol 95.2 N = 0.164 x 6.022 x 1023 = 9.88 x 1022 S: The number of molecules is approximately 9.88 x 1022. 4. Volume of pure ethanol to make 800 mL of 12% (v/v) ethanol solution G: Total volume = 800 ml Concentration = 12% v/v R: Volume of pure ethanol A: Volume of ethanol = = 12 x 800 = 96 mL 100 S: The volume of pure ethanol needed is 96 mL. 5. Maximum mass of fluoride ions in 500.0 mL water if limit is 1.5 ppm GR: number of molecules A: 1. n = M 2. N =n XNA S: 15.6 n = ~0.164 mol 95.2 N = 0.164 x 6.022 x 1025 = 9.88 x 1022 S: The number of molecules is approximately 9.88 x 1022 4. Volume of pure ethanol to make 800 mL of 12% (v/v) ethanol solution G: Total volume = 800 mL Concentration = 12% v/v R: Volume of pure ethanol A Volume of ethanol = - 12 100 X 800 = 96 mL S: The volume of pure ethanol needed is 96 mL. 5. Maximum mass of fluoride ions in 500.0 ml water if limit is 1.5 ppm G. . 1.5 ppm = 1.5 mg/L . Volume = 500.0 ml = 0.500 L R: Maximum mass in mg A: mass = 1.5mg/L x 0.500L = 0.75 mg S: The maximum mass of fluoride ions that can be dissolved is 0.75 mg. 6. Molar concentration of 0.289 mol of FeCls in 120 mL solution G . n = 0.289 mol . V = 120 ml = 0.120 L R: Molar concentration (C) A: n 0.289 C = V 0.120 - 2.41 mol/L S: The molar concentration of FeCly is 2.41 mol/L. 7. Mass of solute in 2.5 L of 1.00 mol/L K2Cr04 G: . C = 1.00 mol/L V = 2.5 L Molar mass of K2Cr04 = 194.2 g/mol R: mass of soluteS: The maximum mass of fluoride ions that can be dissolved is 0.75 mg. 6. Molar concentration of 0.289 mol of FeCls in 120 mL solution G: n = 0.289 mol . V = 120 ml = 0.120 L R: Molar concentration (C) A: C = n 0.289 V 0.120 = 2.41 mol/L S: The molar concentration of FeCls is 2.41 mol/L. 7. Mass of solute in 2.5 L of 1.00 mol/L K2Cr04 G . C = 1.00 mol/L V = 2.5 L . Molar mass of K2Cr04 = 194.2 g/mol R: mass of solute n = C x V = 1.00 x 2.5 = 2.5 mol m = n x M = 2.5 x 194.2 = 485.5 g S: The mass of K2Cr04 solute is 485.5 g. 8. Volume of 1.25 mol/L NaCl to make 50.0 mL of 1.00 mol/L NaCl G: . Ci = 1.25 mol/L . Cf = 1.00 mol/L . Vf = 50.0 mL R: Volume Vi to be used from initial solution A: CiVi = CIVs = Vi = CIVs 1.00 x 50.0 Ci = 40.0 mL 1.25 S: 40.0 mL of 1.25 mol/L NaCl must be used to make the final solution. Question 9 (a): How many moles of Oz are needed to react with 1.2 mol of NH3? Balanced Equation: 2NH3 + 302 + 2CH4 - 2HCN + 6H20 GRASS Method: G Given: 1.2 mol of NH3 . Ratio from equation: 2 mol NH3 : 3 mol Oz R Required: mol of Oz neededS: 40.0 mL of 1.25 mol/L NaCl must be used to make the final solution. Question 9 (a): How many moles of Oz are needed to react with 1.2 mol of NH3? Balanced Equation: 2NH3 + 302 + 2CHA - 2HCN + 6H20 GRASS Method: G Given: 1.2 mol of NH3 Ratio from equation: 2 mol NH3 : 3 mol Oz R . Required: mol of Oz needed A: From ratio: 3 mol O2 2 mol NH3 = x mol O2/1.2 mol NH3 x = , x 1.2 = 1.8 mol Oz S 1.8 moles of oxygen gas (02) are required to react with 1.2 mol of ammonia (NH3). Question 9 (b): How many moles of H2O can be expected from the reaction of 12.5 mol of CH4? Balanced Equation: 2NH3 + 302 + 2CH4 - 2HCN + 6H20 GRASS Method: G Given: 12.5 mol of CH4 Ratio from equation: 2 mol CH4 - 6 mol H20 NH3 and Oz are in excess . Required: mol of H2O formed A: 6 mol H20 From ratio: 2 mol CH4 = x mol H20/12.5 mol CH4 x 12.5 = 3 x 12.5 = 37.5 mol H20 S: 37.5 moles of water (H20) can be expected from the complete reaction of 12.5 mol of methane (CH4). Question 10 Given: Mass of H3PO4 produced = 23.5 g Molar mass of H3PO4 = 97.99 g/mol Balanced equation: PCIs + 4H20 - H3PO4 + 5HCI Required: Mass of PCIs needed (in g)R Required: mol of H2O formed A: From ratio: 6 mol HO 2 mol CH4 = x mol H20/12.5 mol CH4 x 12.5 = 3 x 12.5 = 37.5 mol H20 S: 37.5 moles of water (H20) can be expected from the complete reaction of 12.5 mol of methane (CH4). Question 10 Given: Mass of H3PO4 produced = 23.5 g Molar mass of H3PO4 = 97.99 g/mol Balanced equation: PCls + 4H20 - H3PO4 + 5HCI Required: Mass of PCIs needed (in g) Analysis: 1. Find moles of H3PO4 using: moles = mass / molar mass 2. Use the mole ratio from the balanced equation (1:1 ratio with PCIs) 3. Multiply moles of PCIs by its molar mass (208.22 g/mol) Solution: Moles of H3PO4 = 23.5 g + 97.99 g/mol = 0.240 mol Mole ratio: 1 mol PCIs : 1 mol HaPO4 - 0.240 mol PCIs Mass of PCIs = 0.240 mol x 208.22 g/mol = 49.97 g Sentence: The mass of PCIs needed is approximately 50.0 g. Question 11 Given: Mass of Ba(NO3)2 = 35.0 g Mass of NazSO4 = 25.0 g Balanced equation: Ba(NO3)2 + NazSO4 - BaSO4 + 2NaNO3 Molar masses: Ba(NO3)2 = 261.35 g/mol Na2SO4 = 142.05 g/mol BaSO4 = 233.40 g/mol (a) Limiting Reagent Required: Identify the limiting reagent. Analysis: 1. Convert both masses to moles. 2. Compare moles based on 1:1 ratio from the equation. Solution:Mass of PCIs needed (in g) Analysis: 1. Find moles of H3PO4 using: moles = mass / molar mass 2. Use the mole ratio from the balanced equation (1:1 ratio with PCIs) 3. Multiply moles of PCIs by its molar mass (208.22 g/mol) Solution: Moles of H3PO4 = 23.5 g : 97.99 g/mol = 0.240 mol Mole ratio: 1 mol PCIs : 1 mol H3PO4 - 0.240 mol PCIs Mass of PCIs = 0.240 mol x 208.22 g/mol = 49.97 g Sentence: The mass of PCIs needed is approximately 50.0 g. Question 11 Given: Mass of Ba(NO3)2 = 35.0 g Mass of Na2SO4 = 25.0 g Balanced equation: Ba(NO3)2 + NazSO4 - BaSO4 + 2NaNO3 Molar masses: Ba(NO3)2 = 261.35 g/mol Na2SO4 = 142.05 g/mol BaSO4 = 233.40 g/mol (a) Limiting Reagent Required: Identify the limiting reagent. Analysis: 1. Convert both masses to moles. 2. Compare moles based on 1:1 ratio from the equation. Solution: Moles of Ba(NO3)2 = 35.0 g : 261.35 g/mol = 0.134 mol Moles of Na2SO4 = 25.0 g + 142.05 g/mol = 0.176 mol Since Ba(NO3)2 has fewer moles, it is the limiting reagent. Sentence: The limiting reagent is Ba(NO3)2- (b) Mass of BaSO4 Produced Required: Mass of BaSO4 (in g) Analysis: 1. Use mole ratio: 1 mol Ba(NO3)2 - 1 mol BaSO4 2. Multiply by molar mass of BaSO4 Solution: Moles of BaSO4 = 0.134 mol Mass = 0.134 mol x 233.40 g/mol = 31.3 g Sentence: The mass of BaSO4 that could be produced is 31.3 g.Question 11 Given: Mass of Ba(NO3)2 = 35.0 g Mass of NazSO4 = 25.0 g Balanced equation: Ba(NO3)2 + NazSO4 - BaSO4 + 2NaNO3 Molar masses: Ba(NO3)2 = 261.35 g/mol NazSO4 = 142.05 g/mol BaSO4 = 233.40 g/mol (a) Limiting Reagent Required: Identify the limiting reagent. Analysis: 1. Convert both masses to moles. 2. Compare moles based on 1:1 ratio from the equation. Solution: Moles of Ba(NO3)2 = 35.0 g + 261.35 g/mol = 0.134 mol Moles of Na2SO4 = 25.0 g : 142.05 g/mol = 0.176 mol - Since Ba(NO3)2 has fewer moles, it is the limiting reagent. Sentence: The limiting reagent is Ba(NO3)2. (b) Mass of BaSO4 Produced Required: Mass of BaSO4 (in g) Analysis: 1. Use mole ratio: 1 mol Ba(NO3)2 - 1 mol BaSO4 2. Multiply by molar mass of BaSO4 Solution: Moles of BaSO4 = 0.134 mol Mass = 0.134 mol x 233.40 g/mol = 31.3 g Sentence: The mass of BaSO4 that could be produced is 31.3 g. Q12a. Expected mass of NazCO; produced from 8.0 g of NaHCO3 Given: Mass of NaHCO3 = 8.0 g Balanced equation: 2NaHCO3 - NazCO3 + CO2 + H2O Molar mass of NaHCO3 = 84.0 g/mol Molar mass of NazCO3 = 106.0 g/mol Required: Expected mass of NazCO3 Analysis: . Moles of NaHCO3 = 8.0 g + 84.0 g/mol = 0.0952 mol From the equation: 2 mol NaHCO3 - 1 mol NazCO3Required: Expected mass of NazCO3 Analysis: . Moles of NaHCO3 = 8.0 g + 84.0 g/mol = 0.0952 mol . From the equation: 2 mol NaHCO3 - 1 mol NazCO3 So, moles of NazCO3 = 0.0952 + 2 = 0.0476 mol . Mass of NazCO3 = 0.0476 mol x 106.0 g/mol = 5.05 g Solution: The expected mass of NazCO, is 5.05 g 12(b)(i) Actual yield of NazCO3 Given: . Mass of empty ceramic dish = 9.50 g . Mass of ceramic dish + NaHCO3 before heating = 17.50 g . Mass of ceramic dish after heating = 14.00 g Required: Actual yield of NazCO3 (in grams) Analysis: The actual yield is the mass of solid left after heating, which is calculated by: Mass of NazCO3 = Mass after heating - Mass of empty dish Solution: Mass of NazCO3 = 14.00 g - 9.50 g = 4.50 g Statement: The actual yield of Naz CO, in the experiment is 4.50 g. 12(b) (ii) Percent Yield of NazCO3 Given: . Actual yield of NazCO; = 4.50 g (from part i) . Theoretical (expected) yield of NazCO3 from part (a) = 5.04 g Required: Percent yield of Naz CO3 Analysis: Percent yield is calculated using the formula: Actual Yield Percent Yield = X 100 Theoretical Yield Solution: Percent Yield = 4.50 5.04 x 100 = 89.29% Rounded to two decimal places: 89.29% Statement: The percent yield of NazCO; in the experiment is 89.29%