Quiz 3 - Chemical Calculations | Chemical Calculations

• Sodium sulfate (Na₂SO₄) dissociates into 2 Na⁺ and 1 SO₄^2-, giving 3 ions per formula unit.
• Moles of Na₂SO₄ = 1.20 M x 0.075 L = 0.090 mol.
• Total moles of ions = 0.090 mol x 3 = 0.270 mol.

• Moles of NH₃ = $\frac{3.00}{17.03}$ = 0.176 mol, moles of HCl = $\frac{18.0}{36.46}$ ​= 0.493 mol. NH₃ is the limiting reactant.
• Theoretical yield of NH₄Cl = 0.176 mol x 53.49 g/mol = 9.42 g.

• Total volume = 500.0 mL = 0.500 L
• Moles of NO₃^- from Mg(NO₃)₂ = 0.300 L x 0.200 M x 2= 0.120 mol.
• Moles of NO₃^- required = 0.500 M x 0.500 L = 0.250 mol.
• Moles of NO₃^- from HNO₃ = 0.250 mol - 0.120 mol = 0.130 mol ⇒ Concentration of HNO₃ = $\frac{0.130 \mathrm{mol}}{0.200 \mathrm{L}}$ = 0.650M.

• Dilution equation C₁V₁ = C₂V₂​, where C₁ = 18.0 M, C₂ = 0.55 M, V₂ = 250.0 mL
• V₁ = $\frac{{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1}$ = $\frac{0.55 \mathrm{M} \mathrm{x} 250.0 \mathrm{mL}}{18.0 \mathrm{M}}$ = 7.64 mL

• The balanced reaction is Fe₂S₃ + 6 HCl → 2 FeCl₃ + 3 H₂S.
• Moles of  Fe₂S₃ = $\frac{\mathrm{moles} \mathrm{of} \mathrm{HCl}}{6}$ = $\frac{1.5}{6}$ = 0.25 mol.

• First, calculate the moles of OH^- ions required: moles of OH^- = M x V = 0.100 M x 0.250 L = 0.025 mol
• Since Sr(OH)₂ dissociates completely in water to produce 2 OH^- ions per formula unit: 
  moles of Sr(OH)₂ = $\frac{\mathrm{moles} \mathrm{of} {\mathrm{OH}}^{-}}{2}$ = $\frac{0.025 \mathrm{mol}}{2}$ = 0.0125 mol
• Now, calculate the mass of Sr(OH)₂.8H₂O needed: mass = 0.0125 mol x 265.76 g/mol = 3.32 g

• Molar mass of CaCl₂ =  110.98g/mol
• Calculate moles of CaCl₂: $\frac{7.50 \mathrm{g}}{110.98 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 0.0676 mol
• Calculate concentration of Ca²⁺: $\frac{{\mathrm{n}}_{{\mathrm{CaCl}}_2}}{\mathrm{V}}$ = $\frac{0.0676 \mathrm{mol}}{0.350 \mathrm{L}}$ = 0.193 M

• Calculate the mass of oxygen in the oxide: Mass of O = Mass of Z₂O₃ − Mass of Z = 0.797 g − 0.422 g = 0.375 g
• Calculate the moles of oxygen: moles of O = $\frac{0.375 \mathrm{g}}{16.00 \mathrm{g}.{\mathrm{mol}}^{-1}}$ = 0.02344 mol
• Since the formula of the oxide is Z₂O₃, the moles of Z should be: moles of Z = $\frac{2}{3}$ x moles of O = 0.01563 mol
• Calculate the molar mass of Z: M_Z = $\frac{\mathrm{mass} \mathrm{of} \mathrm{Z}}{\mathrm{moles} \mathrm{of} \mathrm{Z}}$ = $\frac{0.422 \mathrm{g}}{0.01563 \mathrm{mol}}$ = 27.00 g.mol^-1

The element with a molar mass of approximately 27.00 g/mol is aluminum (Al).

• Calculate the mass of I: Mass of KI = 0.300 mol × 166.00 g/mol = 49.80 g; Mass of I in KI = 0.300 mol × 126.90 g/mol = 38.07 g

• Calculate the total mass of the sample: Mass of NaBr = 1.00 mol × 102.89 g/mol = 102.89 g; Total mass = 102.89 g + 49.80 g = 152.69 g

• Calculate the mass percent of iodine: Mass percent of I = (38.07 g / 152.69 g) × 100 ≈ 24.9%

• Calculate moles of Cl⁻ from NaCl:
  Moles of NaCl = 1.50 M × 0.150 L = 0.225 mol. Each mole of NaCl provides 1 mole of Cl⁻ ⇒ Moles of Cl⁻ from NaCl = 0.225 mol
• Calculate moles of Cl⁻ from MgCl₂:
  Moles of MgCl₂ = 0.750 M × 0.250 L = 0.1875 mol. Each mole of MgCl₂ provides 2 moles of Cl⁻ ⇒ Moles of Cl⁻ from MgCl₂ = 0.1875 mol × 2 = 0.375 mol
• Total moles of Cl⁻ = 0.225 mol + 0.375 mol = 0.600 mol
• Total volume = 150 mL + 250 mL = 400 mL = 0.400 L
• Calculate the concentration of Cl^-: Concentration of Cl^- =  $\frac{\mathrm{moles} \mathrm{of} {\mathrm{Cl}}^{-}}{\mathrm{total} \mathrm{volume}}$ = $\frac{0.600 \mathrm{mol}}{0.400 \mathrm{L}}$ = 1.50 M