Question

In the precipitation reaction Na₂(C₂O₄)^2- (aq) + CaCl₂ (aq) -----> Ca(C₂O₄) (s) + 2NaCl (aq), how do I determine how many grams of the reactants I need to yield 5 grams of the precipitate?

Answer 1

Precipitate will be the solid product formed.

Here the solid product formed is Ca(C₂O₄).

Hence, the precipitate formed is Ca(C₂O₄).

Molar mass of Ca(C₂O₄) = 128.1 g/mol

Yield = 5 g

Hence, No.of moles of Ca(C₂O₄) formed = (5 g) / (128.1 g/mol) = 0.039 mol

From the balanced reaction given below:

Na₂(C₂O₄)^2- (aq) + CaCl₂ (aq) -----> Ca(C₂O₄) (s) + 2NaCl (aq)

From 1 mole of Na₂(C₂O₄)^2- and 1 mole of CaCl₂ , 1 mole of Ca(C₂O₄) is formed.

That is :

• 1 mole of Na₂(C₂O₄)^2- -----> 1 mole of Ca(C₂O₄)
• 1 mole of CaCl₂ -----> 1 mole of Ca(C₂O₄)

Therefore, To yield 5 g (0.039 mol) of Ca(C₂O₄):

• 0.039 mol of Ca(C₂O₄) -----> 0.039 mol of Na₂(C₂O₄)^2-
• 0.039 mol of Ca(C₂O₄) -----> 0.039 mol of CaCl₂

Hence, the mass of reactants required:

• Mass of Na₂(C₂O₄)^2- required = (no.of moles) x (molar mass)

= (0.039 mol) x (111 g/mol)

= 4.333 g of Na₂(C₂O₄)^2- is required

• Mass of CaCl₂  required = (no.of moles) x (molar mass)

= (0.039 mol) x (111 g/mol)

= 4.333 g of CaCl₂  is required

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