Chemistry Answer Note #6

Question 1 how many molecules of NH3 are produced by the reaction of 10 moles od Ca(OH)2 according to the following. (NH4)2SO4 + Ca(OH)2 produce NH3 + CaSO4 + H2O Answer In order to answer this question accurately, we need to first understand the equation given. It is a balanced chemical equation, where the reactants (NH4)2SO4 and Ca(OH)2 are reacting to form the products NH3, CaSO4, and H2O. The coefficients in front of each compound indicate the number of molecules involved in the reaction. It is important to note that a mole is a unit of measurement used in chemistry to express the amount of a substance, which is equivalent to Avogadro's number (6.022 x 10^23) of particles. Therefore, 10 moles of Ca(OH)2 means that we have 10 times Avogadro's number of Ca(OH)2 molecules, which is equal to 6.022 x 10^24 molecules. To determine the number of molecules of NH3 produced, we need to use the stoichiometric ratio between Ca(OH)2 and NH3. From the balanced equation, we can see that for every 1 mole of Ca(OH)2, we get 1 mole of NH3. This means that for 10 moles of Ca(OH)2, we will get 10 moles of NH3. Now, to convert moles to molecules, we multiply by Avogadro's number. Therefore, the total number of molecules of NH3 produced by the reaction of 10 moles of Ca(OH)2 is (10 moles)(6.022 x 10^23 molecules/mole) = 6.022 x 10^24 molecules of NH3. Question 2 Ammonia can be generated by heating together ammonium chloride and calcium hydroxide. During the reaction, calcium chloride and water are also formed. If a mixture containing 33.0 g of each NH4Cl and Ca(OH)2 is heated, which reactant remains in excess and in what mass? Answer Based on the given information, we can determine the reactant in excess by using the concept of limiting reagent. This concept states that in a chemical reaction, one reactant will be consumed completely while the other reactant will still be left over. The reactant that is completely consumed is known as the limiting reagent, while the other is in excess. To determine the limiting reagent, we will first need to calculate the moles of each reactant by using their respective molar masses. The molar mass of NH4Cl is 53.49 g/mol and the molar mass of Ca(OH)2 is 74.09 g/mol. Moles of NH4Cl = 33.0 g / 53.49 g/mol = 0.617 mol Moles of Ca(OH)2 = 33.0 g / 74.09 g/mol = 0.445 mol Next, we need to calculate the mole ratio of the reactants based on the balanced chemical equation. NH4Cl + Ca(OH)2 → NH3 + CaCl2 + H2O Moles of NH4Cl : Moles of Ca(OH)2 = 1 : 1 Since the mole ratio of both reactants is the same, the reactant that is present in lesser quantity will be the limiting reagent. From the calculations, we can see that Ca(OH)2 has a lesser number of moles compared to NH4Cl. Therefore, Ca(OH)2 is the limiting reagent and NH4Cl is in excess. To determine the mass of Ca(OH)2 remaining in excess, we will use the following formula: Mass of excess reactant = Initial mass of reactant - Mass of limiting reagent consumed Mass of Ca(OH)2 remaining in excess = 33.0 g - (0.445 mol x 74.09 g/mol) = 2.62 g In conclusion, the reactant in excess is NH4Cl and the mass of Ca(OH)2 remaining in excess is 2.62 g. Question 3 inside the rectangels label each chemical as acid base conjugate acid or conjugate base. Using the behavior of that chemical in teh eaction ot ssign its label. Answer In order to accurately label the chemicals in the reaction as acid, base, conjugate acid, or conjugate base, we must first understand the basic definitions of these terms. An acid is a substance that can donate a proton, while a base is a substance that can accept a proton. A conjugate acid is the species that is formed after a base accepts a proton, and a conjugate base is the species that is formed after an acid donates a proton. Now, let's look at the reaction and assign labels to the chemicals involved. The reaction we are considering is the dissociation of acetic acid (CH3COOH) in water: CH3COOH + H2O ⇌ CH3COO- + H3O+ In this reaction, acetic acid is acting as the acid, as it donates a proton to water, which acts as the base. Therefore, acetic acid can be labeled as the acid and its conjugate base is acetate (CH3COO-). Water, on the other hand, is accepting a proton from acetic acid and therefore acting as the base. Its conjugate acid is hydronium (H3O+), which is formed when water accepts a proton. Question 4 Match the following aqueous solutions with the appropriate letter from the column on the right. fill in the blank 1 1. 9.8×10-2 m Fe2(SO4)3 A. Lowest freezing point fill in the blank 2 2. 0.14 m MnCl2 B. Second lowest freezing point fill in the blank 3 3. 0.13 m Ba(OH)2 C. Third lowest freezing point fill in the blank 4 4. 0.46 m Urea(nonelectrolyte) D. Highest freezing point Answer According to the colligative properties of solutions, the freezing point of a solution decreases as the concentration of solutes increases. This is because the solutes disrupt the orderly arrangement of water molecules, making it harder for them to form a solid structure. Based on this concept, the solutions can be organized from lowest freezing point to highest freezing point as follows: 1. 9.8×10-2 m Fe2(SO4)3 has two moles of ions (Fe3+ and 3SO4 2-) for every mole of compound dissolved. This results in a van't Hoff factor of 3, meaning that the actual concentration of particles in the solution is 3 times higher than the measured concentration. Therefore, the freezing point of this solution would be the lowest, making it the correct match for option A. Source: Chang, R. (2017). General chemistry: the essential concepts (7th ed.). New York: McGraw-Hill Education. 2. 0.14 m MnCl2 has two moles of ions (Mn2+ and 2Cl-) for every mole of compound dissolved. This results in a van't Hoff factor of 3, meaning that the actual concentration of particles in the solution is 3 times higher than the measured concentration. Since this is a higher concentration than 9.8×10-2 m Fe2(SO4)3, it will have a slightly higher freezing point and can be matched with option B. Source: Petrucci, R.H., Herring, F.G., Madura, J.D., & Bissonnette, C. (2016). General chemistry: principles and modern applications (11th ed.). New York: Pearson. 3. 0.13 m Ba(OH)2 has three moles of ions (Ba2+ and 2OH-) for every mole of compound dissolved. This results in a van't Hoff factor of 3, making the actual concentration of particles in the solution equal to the measured concentration. Since this is a lower concentration than 0.14 m MnCl2, it will have a slightly lower freezing point and can be matched with option C. Source: Brown, T.L., Lemay, H.E., Bursten, B.E., Murphy, C., Woodward, P., & Stoltzfus, M.E. (2017). Chemistry: the central science (14th ed.). New York: Pearson. 4. 0.46 m Urea is a nonelectrolyte solution, meaning it does not dissociate into ions when dissolved in water. This results in a van't Hoff factor of 1 and does not have an effect on the freezing point. Therefore, the correct match for this solution would be option D, highest freezing point. Source: Tro, N.J. (2017). Chemistry: a molecular approach (4th ed.). New York: Pearson. Question 5 The boiling point of ethanol, , is 78.500 °C at 1 atmosphere. In a laboratory experiment, students synthesized a new compound and found that when 13.24 grams of the compound were dissolved in 259.4 grams of ethanol, the solution began to boil at 78.716 °C. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound? Answer Based on the given information, the students were able to determine the boiling point elevation of the ethanol solution due to the presence of the new compound. This phenomenon, known as boiling point elevation, is a colligative property which is dependent on the number of solute particles present in a solution. The higher the number of solute particles, the greater the boiling point elevation. To determine the molecular weight of the compound, we can use the equation: ∆Tb = Kb * m * i, where ∆Tb is the boiling point elevation, Kb is the molal boiling point elevation constant, m is the molality of the solution, and i is the van't Hoff factor. In this case, we can assume that the van't Hoff factor is equal to 1 since the compound is non-electrolyte. Using the known values, we can rearrange the equation to solve for the molecular weight (M): M = ∆Tb / (Kb * m) = (78.716 °C - 78.500 °C) / (0.51 molal^-1 * 0.0515 mol / kg) = 0.216 °C / 0.026315 mol = 8.219 g/mol Therefore, the molecular weight of the new compound is 8.219 g/mol. Step-by-step breakdown: 1. Calculate the molality (m) of the solution using the formula: m = moles of solute / mass of solvent in kg. m = 13.24 g compound / (259.4 g ethanol / 1000 g / kg) = 0.0515 mol / kg 2. Substitute the given values in the equation: ∆Tb = Kb * m * i and solve for the molal boiling point elevation constant (Kb). Kb = ∆Tb / (m * i) = (78.716 °C - 78.500 °C) / (0.0515 mol / kg * 1) = 0.51 molal^-1 3. Substitute the known values in the equation: M = ∆Tb / (Kb * m) and solve for the molecular weight (M). M = (78.716 °C - 78.500 °C) / (0.51 molal^-1 * 0.0515 mol / kg) = 0.216 °C / 0.026315 mol = 8.219 g/mol