Answers – Moles

Answer:

a) Number of moles of magnesium (Mg):

• Molar mass of Mg = 24 g/mol
• Number of moles = Mass ÷ Molar mass
• Number of moles = 24.0 g ÷ 24 g/mol = 1.0 mol

b) Number of moles of carbon (C):

• Molar mass of C = 12 g/mol
• Number of moles = Mass ÷ Molar mass
• Number of moles = 6.0 g ÷ 12 g/mol = 0.5 mol

c) Number of moles of sodium hydroxide (NaOH):

• Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
• Number of moles = Mass ÷ Molar mass
• Number of moles = 4.0 g ÷ 40 g/mol = 0.1 mol

Answer:

a) Mass of 0.5 moles of oxygen gas (O₂):

• Molar mass of O₂ = 16 × 2 = 32 g/mol
• Mass = Number of moles × Molar mass
• Mass = 0.5 mol × 32 g/mol = 16 g

b) Mass of 0.25 moles of calcium carbonate (CaCO₃):

• Molar mass of CaCO₃ = 40 + 12 + (16 × 3) = 40 + 12 + 48 = 100 g/mol
• Mass = Number of moles × Molar mass
• Mass = 0.25 mol × 100 g/mol = 25 g

c) Mass of 0.01 moles of potassium permanganate (KMnO₄):

• Molar mass of KMnO₄ = 39 + 55 + (16 × 4) = 39 + 55 + 64 = 158 g/mol
• Mass = Number of moles × Molar mass
• Mass = 0.01 mol × 158 g/mol = 1.58 g

Answer:

a) Volume of 0.75 moles of hydrogen gas (H₂) at RTP:

• At room temperature and pressure, 1 mole of any gas occupies 24 dm³
• Volume = Number of moles × 24 dm³/mol
• Volume = 0.75 mol × 24 dm³/mol = 18 dm³

b) Volume of 0.125 moles of carbon dioxide (CO₂) at RTP:

• Volume = Number of moles × 24 dm³/mol
• Volume = 0.125 mol × 24 dm³/mol = 3 dm³

Answer:

a) Percentage by mass of hydrogen in ammonia (NH₃):

• Molar mass of NH₃ = 14 + (3 × 1) = 14 + 3 = 17 g/mol
• Mass of hydrogen in NH₃ = 3 × 1 = 3 g
• Percentage by mass of hydrogen = (Mass of hydrogen ÷ Molar mass of NH₃) × 100%
• Percentage by mass of hydrogen = (3 ÷ 17) × 100% = 17.65%

b) Percentage by mass of oxygen in calcium carbonate (CaCO₃):

• Molar mass of CaCO₃ = 40 + 12 + (3 × 16) = 40 + 12 + 48 = 100 g/mol
• Mass of oxygen in CaCO₃ = 3 × 16 = 48 g
• Percentage by mass of oxygen = (Mass of oxygen ÷ Molar mass of CaCO₃) × 100%
• Percentage by mass of oxygen = (48 ÷ 100) × 100% = 48%

Answer:

a) Mass of zinc needed to produce 12 dm³ of hydrogen gas at RTP:

• Step 1: Calculate the number of moles of hydrogen
• At RTP, 1 mole of gas occupies 24 dm³
• Number of moles of H₂ = Volume ÷ 24 = 12 ÷ 24 = 0.5 mol
• Step 2: Use the equation to find moles of zinc needed
• From the equation: Zn + 2HCl → ZnCl₂ + H₂
• 1 mol of Zn produces 1 mol of H₂
• Therefore, 0.5 mol of H₂ requires 0.5 mol of Zn
• Step 3: Calculate the mass of zinc
• Molar mass of Zn = 65 g/mol
• Mass of Zn = Number of moles × Molar mass = 0.5 mol × 65 g/mol = 32.5 g

b) Volume of hydrogen gas produced when 6.5g of zinc fully reacts with excess acid:

• Step 1: Calculate the number of moles of zinc
• Molar mass of Zn = 65 g/mol
• Number of moles of Zn = Mass ÷ Molar mass = 6.5 g ÷ 65 g/mol = 0.1 mol
• Step 2: Use the equation to find moles of hydrogen produced
• From the equation: 1 mol of Zn produces 1 mol of H₂
• Therefore, 0.1 mol of Zn produces 0.1 mol of H₂
• Step 3: Calculate the volume of hydrogen
• At RTP, 1 mole of gas occupies 24 dm³
• Volume of H₂ = Number of moles × 24 = 0.1 mol × 24 dm³/mol = 2.4 dm³

Answer:

a) Number of moles of aluminum:

• Molar mass of Al = 27 g/mol
• Number of moles of Al = Mass ÷ Molar mass = 9.0 g ÷ 27 g/mol = 0.333 mol

b) Number of moles of oxygen:

• Molar mass of O₂ = 32 g/mol
• Number of moles of O₂ = Mass ÷ Molar mass = 9.6 g ÷ 32 g/mol = 0.3 mol

c) Identifying the limiting reactant:

• From the balanced equation: 4Al + 3O₂ → 2Al₂O₃
• The ratio is 4 mol Al : 3 mol O₂
• For 0.333 mol Al, we need (3 ÷ 4) × 0.333 = 0.25 mol O₂
• We have 0.3 mol O₂, which is more than needed
• Therefore, aluminum is the limiting reactant

d) Mass of aluminum oxide produced:

• From the equation: 4 mol Al produces 2 mol Al₂O₃
• So 0.333 mol Al will produce (2 ÷ 4) × 0.333 = 0.1665 mol Al₂O₃
• Molar mass of Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102 g/mol
• Mass of Al₂O₃ = Number of moles × Molar mass = 0.1665 mol × 102 g/mol = 16.98 g or approximately 17.0 g

Answer:

a) Balanced equation for the decomposition of copper carbonate:

CuCO₃ → CuO + CO₂

b) Calculate the mass of pure copper carbonate in the original sample:

• Step 1: Find moles of CuO produced
• Molar mass of CuO = 63.5 + 16 = 79.5 g/mol
• Moles of CuO = Mass ÷ Molar mass = 0.16 g ÷ 79.5 g/mol = 0.00201 mol
• Step 2: Using the equation, find moles of CuCO₃
• From the equation: 1 mol CuCO₃ produces 1 mol CuO
• Therefore, moles of CuCO₃ = 0.00201 mol
• Step 3: Calculate mass of pure CuCO₃
• Molar mass of CuCO₃ = 63.5 + 12 + (3 × 16) = 63.5 + 12 + 48 = 123.5 g/mol
• Mass of pure CuCO₃ = Number of moles × Molar mass = 0.00201 mol × 123.5 g/mol = 0.248 g

c) Calculate the percentage purity of the original copper carbonate sample:

• Percentage purity = (Mass of pure CuCO₃ ÷ Mass of impure sample) × 100%
• Percentage purity = (0.248 g ÷ 0.25 g) × 100% = 99.2%

Answer:

To find which two samples contain the same number of atoms, I’ll calculate the number of atoms in each:

a) 2.0g of hydrogen gas (H₂):

• Molar mass of H₂ = 2 g/mol
• Number of moles = 2.0 g ÷ 2 g/mol = 1.0 mol
• Each molecule of H₂ contains 2 hydrogen atoms
• Number of atoms = 1.0 mol × 6.02 × 10²³ molecules/mol × 2 atoms/molecule = 12.04 × 10²³ atoms

b) 16.0g of oxygen gas (O₂):

• Molar mass of O₂ = 32 g/mol
• Number of moles = 16.0 g ÷ 32 g/mol = 0.5 mol
• Each molecule of O₂ contains 2 oxygen atoms
• Number of atoms = 0.5 mol × 6.02 × 10²³ molecules/mol × 2 atoms/molecule = 6.02 × 10²³ atoms

c) 14.0g of nitrogen gas (N₂):

• Molar mass of N₂ = 28 g/mol
• Number of moles = 14.0 g ÷ 28 g/mol = 0.5 mol
• Each molecule of N₂ contains 2 nitrogen atoms
• Number of atoms = 0.5 mol × 6.02 × 10²³ molecules/mol × 2 atoms/molecule = 6.02 × 10²³ atoms

d) 4.0g of helium (He):

• Molar mass of He = 4 g/mol
• Number of moles = 4.0 g ÷ 4 g/mol = 1.0 mol
• Each atom of He is already an atom (not a molecule)
• Number of atoms = 1.0 mol × 6.02 × 10²³ atoms/mol = 6.02 × 10²³ atoms

From the calculations, we can see that 16.0g of oxygen gas (O₂), 14.0g of nitrogen gas (N₂), and 4.0g of helium (He) all contain the same number of atoms: 6.02 × 10²³ atoms.

Answer:

To calculate the volume of sulfur dioxide gas produced:

• Step 1: Calculate the number of moles of sodium thiosulfate
• Molar mass of Na₂S₂O₃ = (2 × 23) + (2 × 32) + (3 × 16) = 46 + 64 + 48 = 158 g/mol
• Number of moles = Mass ÷ Molar mass = 12.4 g ÷ 158 g/mol = 0.0785 mol
• Step 2: Use the equation to find moles of sulfur dioxide produced
• From the equation: 1 mol Na₂S₂O₃ produces 1 mol SO₂
• Therefore, moles of SO₂ = 0.0785 mol
• Step 3: Calculate the volume of sulfur dioxide
• At RTP, 1 mole of gas occupies 24 dm³
• Volume of SO₂ = Number of moles × 24 = 0.0785 mol × 24 dm³/mol = 1.88 dm³

The volume of sulfur dioxide gas produced would be 1.88 dm³ at RTP.