Question

1. Calculate the number of moles of calcium chloride that would be necessary to prepare 85.0 g of calcium phosphate

2. Calculate the number of moles of barium sulfate that can be prepared from 60.0 g of barium chloride

3. Calculate the number of grams of zinc chloride that can be prepared from 34.0 g of zinc reacting with hydrochloric acid

Answer 1

1) Molar mass of calcium phosphate Ca₃(PO₄)₂ is 310.17 g/mol

so 85.0 g => 85.0 g / 310.17 g/mol ==> 0.274 mol

1 mol of Ca₃(PO₄)₂ contains 3 mol of Ca²⁺ ions   but 1 mol of CaCl₂ calcium chloride contains 1 mol of Ca²⁺ ions

So number of moles of calcium chloride required is 3 x 0.274 mol => 0.822 moles.

2) 1 mol BaCl₂ contains 1 mol of Ba²⁺ ions and also 1 mol of BaSO₄ contains 1 mol of Ba²⁺ ions

60 g of BaCl₂ = 60 g / 208.23 g/mol ==> 0.2881 mol

moles of BaCl₂ = moles of BaSO₄ ==> 0.2881 moles

3) molar mass of Zn = 65.38 g/mol

So 34.0 g => 34.0 g / 65.38 g/mol = 0.5200 mol

Hence 0.5200 mol of ZnCl₂ can be prepared

Molar mass of ZnCl₂ = 136.28 g/mol

Therefore 0.5200 mol of ZnCl₂ x 136.28 g/mol = 70.87 g of ZnCl₂