Identify the limited reactant and show the work for the following?

(a)
Molar mass of NH₄OH = (14.01 + 1.008×5 + 16.00) g/mol = 35.05 g/mol
Molar mass of HCl = (1.008 + 35.45 g/mol) = 36.46 g/mol

NH₄OH + HCl → NH₄Cl + H₂O
Mole ratio NH₄OH : HCl = 1 : 1

Initial number of moles of NH₄OH = (10.000 g) / (35.05 g/mol) = 0.2853 mol
Initial number of moles of HCl = (10.500 g) / (36.46 g/mol) = 0.2879 > 0.2853 mol

Hence, NH₄OH is the limiting reactant.


====
(b)
(CH₃COO)₂Pb(aq) + 2NaI(aq) → PbI₂(aq) + CH₃COONa(aq)
Mole ratio (CH₃COO)₂Pb : NaI(aq) = 1 : 2

Initial no. of moles of (CH₃COO)₂Pb = (0.500 mmol/mL) × (10.00 mL) = 5.00 mmol
Initial no. of moles of NaI = (0.700 mmol/mL) × (15.00 mL) = 10.5 mmol > 2 × 5.00 mmol

Hence, (CH₃COO)₂Pb is the limiting reactant.


====
(c)
Molar mass of Na₂CO₃ = (22.99×2 + 12.01 + 16.00×3) g/mol = 105.99 g/mol

Na₂CO₃(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l) + CO₂(g)
Mole ratio Na₂CO₃ : H₂SO₄ = 1 : 1

Initial no. of moles of Na₂CO₃ = (5.000 g) / (105.99 g/mol) = 0.04717 mol
Initial no. of moles of H₂SO₄ = (1.000 mol/L) × (45.00/1000) = 0.04500 mol < 0.04717 mol

Hence, H₂SO₄ is the limiting reactant.


====
(d)
2AgNO₃(aq) + Na₂S(aq) → Ag₂S(s) + 2NANO₃(aq)
Mole ratio AgNO₃ : Na₂S = 2 : 1

Initial no. of moles of AgNO₃ = (2.50 mmol/mL) × (35.00 mL) = 87.5 mmol
Initial no. of moles of Na₂S = (0.350 mmol/mL) × (20.00 mL) = 7 mmol < (1/2) × 87.5 mmol

Hence, Na₂S is the limiting reactant.