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1.
a.
Molar mass of BaCl₂•2H₂O = (137.3 + 35.5×2 + 1.0×4 + 16.0×2) g/mol = 244.3 g/mol
Molar mass of BaCl₂•H₂O = (137.3 + 35.5×2 + 1.0×2 + 16.0) g/mol = 226.3 g/mol

BaCl₂•2H₂O(s) → BaCl₂•H₂O(s) + H₂O(g)
Mole ratio BaCl₂•2H₂O : BaCl₂•H₂O = 1 : 1

No. of moles of BaCl₂•2H₂O used = (1.45 g) / (244.3 g/mol) = 0.005935 mol
Maximum no. of moles of BaCl₂•H₂O produced = 0.005935 mol
Theoretical yield of BaCl₂•H₂O = (0.005935 mol) × (226.3 g/mol) = 1.34 g (to 3 sig. fig.)

b.
Molar mass of BaCl₂ = (137.3 + 35.5×2) g/mol = 208.3 g/mol

BaCl₂•2H₂O(s) → BaCl₂(s) + H₂O(g)
Mole ratio BaCl₂•2H₂O : BaCl₂ = 1 : 1

No. of moles of BaCl₂•2H₂O used = 0.005935 mol
Maximum no. of moles of BaCl₂ produced = 0.005935 mol
Theoretical yield of BaCl₂ = (0.005935 mol) × (208.3 g/mol) = 1.24 g (to 3 sig. fig.)

c.
Molar mass of BaO = (137.3 + 16.0) g/mol = 153.3 g/mol

BaCl₂•2H₂O(s) → BaO(s) + H₂O(g) + 2HCl(g)
Mole ratio BaCl₂•2H₂O : BaO = 1 : 1

No. of moles of BaCl₂•2H₂O used = 0.005935 mol
Maximum no. of moles of BaO produced = 0.005935 mol
Theoretical yield of BaO = (0.005935 mol) × (153.3 g/mol) = 0.910 g (to 3 sig. fig.)