Question

How many grams of calcium hydroxide must react to give 3.09 g of Ca3(PO4)2?
Na3PO4(aq)+Ca(OH)2(aq)→Ca3(PO4)2(s)+NaOH(aq)

Answer 1

Molar mass:

Ca(OH)₂ = 74.093 g/mol
Ca₃(PO₄)₂ = 310.17 g/mol

Mole ratio:

2 Na₃PO₄ + 3 Ca(OH)₂ = Ca₃(PO₄)₂ + 6 NaOH

3 x 74.093 g Ca(OH)₂ ----------- 310.17 g Ca₃(PO₄)₂
            ?? g Ca(OH)₂ ----------- 3.09 g Ca₃(PO₄)₂

Mass of Ca(OH)₂ = 3.09 x 3 x 74.093 / 310.17

Mas of Ca(OH)₂ = 686.84211 / 310.17

= 2.214 g of Ca(OH)₂

hope this helps!

Answer 2

The mass of calcium hydroxide is 2.214 grams this can be calculated by using stoichiometric ratio in the given reaction.

According to stoichiometric ratio:

2 Na₃PO₄ + 3 Ca(OH)₂ → Ca₃(PO₄)₂ + 6 NaOH

What information do we have?

Molar mass of Ca(OH)₂ = 74.093 g/mol

Molar mass of Ca₃(PO₄)₂ = 310.17 g/mol

3 x 74.093 g Ca(OH)₂ → 310.17 g Ca₃(PO₄)₂

         ?? g Ca(OH)₂ → 3.09 g Ca₃(PO₄)₂

Calculation for mass:

Mass of Ca(OH)₂ = 3.09 x 3 x 74.093 / 310.17

Mass of Ca(OH)₂ = 686.84211 / 310.17

Mass of Ca(OH)₂= 2.214 g

Find more information about molar mass here:

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