1.9 - Practice Problem Solutions (Outline)

To calculate the number of moles in 25.0 g of calcium carbonate (CaCO₃), we first need to find its molar mass. The molar mass is the sum of the atomic masses of all atoms present in the compound: calcium (Ca) is approximately 40.08 g/mol, carbon (C) is about 12.01 g/mol, and oxygen (O) is roughly 16.00 g/mol. Since there are three oxygen atoms in CaCO₃, we multiply 16.00 by 3. Adding these values together gives us the molar mass of CaCO₃ as 100.09 g/mol. Then, to find the number of moles, we divide the mass (25.0 g) by the molar mass (100.09 g/mol), resulting in approximately 0.2498 moles.

To find the number of formula units of CaCO₃ corresponding to 0.2498 moles, we use Avogadro's number, which tells us how many entities are in one mole of any substance (approximately 6.022 × 10²³). We multiply the number of moles (0.2498) by Avogadro's number, resulting in about 1.505 × 10²³ formula units of CaCO₃.

To determine the mass from a known number of formula units (1.50 × 10²³), we first convert formula units back to moles using Avogadro's number: we divide 1.50 × 10²³ by 6.022 × 10²³, resulting in approximately 0.2492 moles. Next, we find the mass by multiplying the number of moles by the molar mass of CaCO₃ (100.09 g/mol). This calculation yields a mass of about 24.93 g.

Balancing a chemical equation is crucial because it ensures the law of conservation of mass is observed; atoms cannot be lost or gained in a reaction. We start with the unbalanced equation and analyze the number of atoms for each element on both sides. By iteratively adding coefficients and ensuring each element's number matches on both sides, we arrive at the balanced equation: Al₂S₃ + 6 H₂O ⟶ 2 Al(OH)₃ + 3 H₂S.

To calculate the mass of hydrogen sulfide (H₂S) produced from the decomposition of aluminum sulfide (Al₂S₃), we first need the molar mass of Al₂S₃. After calculating, we find that 12.0 g of Al₂S₃ corresponds to approximately 0.07991 moles. From the balanced equation, it produces 3 moles of H₂S for every mole of Al₂S₃ decomposed. Therefore, for 0.07991 moles of Al₂S₃, we can produce 0.2397 moles of H₂S. Finally, we convert this into grams by multiplying the number of moles by the molar mass of H₂S (34.076 g/mol), resulting in approximately 8.17 g of H₂S.

In this problem, we start by calculating the molar mass of dinitrogen pentoxide (N₂O₅), which is essential to find out how many grams can be produced from its decomposition. The balanced equation shows that when 2 moles of N₂O₅ decompose, they produce 4 moles of nitrogen dioxide (NO₂) and 1 mole of oxygen gas (O₂). By knowing we have 1.20 moles of N₂O₅, we can use stoichiometry to find we will get 2.40 moles of NO₂ and 0.600 moles of O₂. Each of these amounts can then be converted into grams using their respective molar masses (NO₂ and O₂) to find the total mass produced.

To calculate the percent yield, we compare the actual yield of the product obtained (120.0 g of NO₂) to the theoretical yield calculated earlier (110.4 g). Using the formula for percent yield: (actual yield / theoretical yield) × 100 %, we find that the percent yield is 108.7%. A yield greater than 100% suggests there may have been an error in the experiment, possibly from impurities in the product or an inaccurate measurement.

To prepare a solution of potassium chromate (K₂CrO₄) with a certain molarity, we start by determining the molar mass. After calculating, we find the molar mass to be 194.20 g/mol. Next, we need 0.200 moles per liter in 0.500 liters, which amounts to 0.100 moles. Converting moles to grams gives us 19.42 g of K₂CrO₄, which we then measure out. The procedure involves adding the solid to a volumetric flask, dissolving it in some distilled water, and then filling it to the final mark with water to create a precise solution.

In this problem, we are diluting an initial solution of potassium chromate. When diluting, we use the dilution formula C1V1=C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution used, C2 is the final concentration after dilution, and V2 is the final volume. Here, we start with an initial concentration of 0.200 M and a volume of 50.0 mL, and we want to know the concentration after diluting it to 250 mL, resulting in a final concentration of 0.0400 M.

In this problem, we determine the molarity of a sodium hydroxide (NaOH) solution. We start by calculating the molar mass of NaOH to be about 40.00 g/mol. Given 8.00 g of NaOH, we find the number of moles by dividing the mass by the molar mass, resulting in 0.200 moles. Since the total volume of the solution is 250.0 mL (equivalent to 0.250 L), we calculate the molarity with the formula M = moles/volume, yielding a solution molarity of 0.800 M.

To find the molality of the sodium hydroxide (NaOH) solution, we first need to determine the total mass of the solution, using its density (1.07 g/mL) and the volume (250.0 mL), which gives us a mass of 267.5 g. We then subtract the mass of the solute (8.00 g of NaOH) from the total mass to find the mass of the solvent (water), which is approximately 259.5 g or 0.2595 kg. Finally, we divide the moles of solute (0.200 mol) by the mass of the solvent in kilograms to calculate the molality, resulting in around 0.771 m.