- The balanced equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. Moles of NaOH = 0.0432 L × 0.236 mol/L = 0.0101952 mol. From the stoichiometry, moles of H₂SO₄ = 0.0101952 mol / 2 = 0.0050976 mol. Concentration of H₂SO₄ = 0.0050976 mol / 0.0367 L ≈ 0.139 M.
- The balanced equation is NaOH + HCl → H₂O + NaCl. Moles of NaOH = 0.0432 L × 0.236 mol/L = 0.0101952 mol. From the stoichiometry, moles of HCl = 0.0101952 mol. Concentration of HCl = 0.0101952 mol / 0.0367 L ≈ 0.278 M.
- The balanced equation is Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O. Molar mass of Al(OH)₃ = 78.00 g/mol. Moles of Al(OH)₃ = 5.8 g / 78.00 g/mol ≈ 0.07436 mol. Moles of HCl required = 3 × 0.07436 mol ≈ 0.22308 mol. Volume of 1.2 M HCl = 0.22308 mol / 1.2 mol/L ≈ 0.1859 L = 186 mL.
- The balanced equation is H₂SO₄ + Na₂CO₃ → Na₂SO₄ + H₂O + CO₂. Molar mass of Na₂CO₃ = 105.99 g/mol. Moles of Na₂CO₃ = 0.364 g / 105.99 g/mol ≈ 0.003434 mol. From the stoichiometry, moles of H₂SO₄ = 0.003434 mol. Concentration of H₂SO₄ = 0.003434 mol / 0.0400 L ≈ 0.0858 M.
- The balanced equation is Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. Molar mass of Ca(OH)₂ = 74.09 g/mol. Moles of Ca(OH)₂ = 5.00 g / 74.09 g/mol ≈ 0.06749 mol. Moles of HCl required = 2 × 0.06749 mol ≈ 0.13498 mol. Volume of 0.100 M HCl = 0.13498 mol / 0.100 mol/L ≈ 1.35 L.
- The balanced equation is Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O. Moles of Ca(OH)₂ = 15.0 g / 74.09 g/mol ≈ 0.2025 mol. Moles of HCl = 0.0750 L × 0.500 mol/L = 0.0375 mol. HCl is limiting (requires 2×0.2025 = 0.405 mol HCl for all Ca(OH)₂). Moles of CaCl₂ formed = 0.0375 mol HCl / 2 = 0.01875 mol. Molar mass of CaCl₂ = 110.98 g/mol. Mass of CaCl₂ = 0.01875 mol × 110.98 g/mol ≈ 2.08 g.
Parent Tip: Review the logic above to help your child master the concept of molarity problems worksheet.
