Detailed Explanation

In Problem 1, students will work with the concept of moles and how they relate to mass and particles.

1. (a) To find the number of moles in a substance, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

First, we need the molar mass of calcium carbonate (CaCO₃):
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (there are 3 oxygen atoms in CaCO₃)

Therefore, the molar mass of CaCO₃ = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol. Now we can calculate the number of moles:

\[ \text{Number of moles} = \frac{25.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.2498 \, \text{mol} \]

1. (b) To find the number of formula units, we use Avogadro's number (6.022 × 10²³). The relationship is:

\[ \text{Number of formula units} = \text{number of moles} \times 6.022 \times 10^{23} \]

So:
\[ \text{Number of formula units} = 0.2498 \, \text{mol} \times 6.022 \times 10^{23} = 1.505 \times 10^{23} \text{ formula units} \]

1. (c) To find the mass of a specific number of formula units, first convert the formula units to moles and then to grams:

\[ \text{number of moles} = \frac{1.50 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.2492 \, \text{mol} \]
\[ \text{mass} = 0.2492 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 24.93 \, \text{g} \]

Detailed Explanation

This problem is focused on balancing chemical equations and performing stoichiometric calculations.

1. (a) To balance the equation, we need to ensure that the number of atoms of each element is equal on both sides. Start with the unbalanced equation:
   \[ \text{Al}_2\text{S}_3 + \text{H}_2\text{O} \longrightarrow \text{Al(OH)}_3 + \text{H}_2 ext{S} \]
   Count the atoms on each side:
2. Aluminum (Al): 2 on left, 1 on right.
3. Sulfur (S): 3 on left, 1 on right.
4. Hydrogen (H) and Oxygen (O) also need to be balanced.

By adjusting coefficients systematically, we can arrive at the balanced equation:
\[ \text{Al}_2\text{S}_3 + 6 \text{H}_2\text{O} \longrightarrow 2 \text{Al(OH)}_3 + 3 \text{H}_2 ext{S} \]

1. (b) To find the mass of H₂S produced, we start by converting 12.0 g of Al₂S₃ into moles. The molar mass of Al₂S₃ is 150.14 g/mol:
   \[ \text{Moles of Al}_2\text{S}_3 = \frac{12.0 \, \text{g}}{150.14 \, \text{g/mol}} \approx 0.07991 \, \text{mol} \]
   Since the balanced equation shows that 1 mol of Al₂S₃ produces 3 mol of H₂S, we find moles of H₂S:
   \[ \text{Moles of H}_2 ext{S} = 0.07991 \, \text{mol Al}_2\text{S}_3 \times 3 = 0.2397 \, \text{mol H}_2 ext{S} \]

Finally, converting moles back to grams using the molar mass of H₂S (34.08 g/mol):
\[ \text{Mass of H}_2 ext{S} = 0.2397 \, \text{mol} \times 34.08 \, \text{g/mol} \approx 8.17 \, \text{g} \]

Detailed Explanation

This problem relates to the concepts of limiting reagents and the percent yield of a reaction.

1. (a) Given the balanced reaction
   \[ 2 ext{N}_2 ext{O}_5(g) \longrightarrow 4 ext{NO}_2(g) + ext{O}_2(g) \]
   We need to determine how much NO₂ and O₂ is generated from 1.20 mol of N₂O₅. Using mole ratios from the balanced equation:
2. For NO₂: \( 1.20 \, \text{mol N}_2 ext{O}_5 \times \frac{4 \, \text{mol NO}_2}{2 \, \text{mol N}_2 ext{O}_5} = 2.40 \, \text{mol NO}_2 \)
3. For O₂: \( 1.20 \, \text{mol N}_2 ext{O}_5 \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol N}_2 ext{O}_5} = 0.600 \, \text{mol O}_2 \)
   Now to find the mass of each gas, we can use their molar masses (NO₂: 46.01 g/mol; O₂: 32.00 g/mol):
   \[ \text{Mass of NO}_2 = 2.40 \, \text{mol} \times 46.01 \, \text{g/mol} = 110.4 \, \text{g} \]
   \[ \text{Mass of O}_2 = 0.600 \, \text{mol} \times 32.00 \, \text{g/mol} = 19.2 \, \text{g} \]
4. (b) To calculate the percent yield, we compare the actual yield to the theoretical yield. Here, the actual yield is given as 120.0 g:
   \[ \text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% = \left( \frac{120.0 \, \text{g}}{110.4 \, \text{g}} \right) \times 100\% = 108.7\% \]
   Since the percent yield is greater than 100%, it indicates a possible error in measurement or product purity.

Detailed Explanation

This problem focuses on preparing solutions and understanding dilutions.

1. (a) To prepare a specific molarity:
   First, calculate the molar mass of potassium chromate (K₂CrO₄):
2. Potassium (K): 39.10 g/mol (there are 2 potassium atoms)
3. Chromium (Cr): 52.00 g/mol
4. Oxygen (O): 16.00 g/mol (4 oxygen atoms)

Molar mass = 2(39.10) + 52.00 + 4(16.00) = 194.20 g/mol.

To find the mass for a 0.200 M solution in 0.500 L:
\[ \text{Required moles} = 0.200 \, \text{mol/L} \times 0.500 \, \text{L} = 0.100 \, \text{mol} \]
\[ \text{Mass} = 0.100 \, \text{mol} \times 194.20 \, \text{g/mol} = 19.42 \, g \]

The procedure involves weighing this mass, dissolving in water in a volumetric flask, and bringing the total volume up to 500 mL.

1. (b) For dilutions, use the dilution equation:
   \[ C₁ V₁ = C₂ V₂ \]
   Where:
2. C₁ = initial concentration = 0.200 M
3. V₁ = initial volume = 50.0 mL = 0.0500 L
4. V₂ = final volume = 250 mL = 0.250 L

Rearranging gives:
\[ C₂ = \frac{C₁ V₁}{V₂} = \frac{0.200 \, \text{mol/L} \times 0.0500 \, ext{L}}{0.250 \, ext{L}} = 0.0400 \, ext{mol/L} \]

Detailed Explanation

This problem involves calculating both molarity and molality of a solution.

1. (a) To find molarity (M), we use:
   \[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
   First, calculate the molar mass of NaOH:
2. Sodium (Na): 22.99 g/mol
3. Oxygen (O): 16.00 g/mol
4. Hydrogen (H): 1.008 g/mol

So, the molar mass = 22.99 + 16.00 + 1.008 = 40.00 g/mol.

Now, find the number of moles:
\[ \text{Moles of NaOH} = \frac{8.00 \, \text{g}}{40.00 \, \text{g/mol}} = 0.200 \, \text{mol} \]

Finally, convert volume from mL to L:
250.0 mL = 0.250 L.

Molarity:
\[ M = \frac{0.200 \, \text{mol}}{0.250 \, \text{L}} = 0.800 \, ext{M} \]

1. (b) To calculate molality (m), we need the mass of the solvent. First, find the total mass of the solution using density:
   \[ \text{Density} = \frac{\text{mass}}{\text{volume}} \implies \text{Mass} = \text{Density} \times \text{Volume} = 1.07 \,\text{g/mL} \times 250.0 \, \text{mL} = 267.5 \, \text{g} \]

The mass of the solvent (water):
\[ \text{Mass of solvent} = 267.5 \, \text{g} - 8.00 \, \text{g} = 259.5 \, \text{g} \approx 0.2595 \, \text{kg} \]

Then calculate molality:
\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.200 \, \text{mol}}{0.2595 \, \text{kg}} \approx 0.771 \, m \]