Question

what is the mass of calcium phosphate that can be prepared from 1.78g of Na3PO4?
Na3PO4(aq) + Ca(OH)2 (aq)---> Ca3(PO4)(s)+NaOH(aq)

Answer 1

The balanced equation is as follows:

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

Given that

1.78 g of Na3PO4

Number of moles = amount in g / molar mass

= 1.78 g of Na3PO4/ 163.94 g/mol

= 0.0109 moles Na3PO4

Now calculate the mole of calcium phosphate as follows:

0.0109 moles Na3PO4* 1 mole Ca3(PO4)2 /2 mole Na3PO4

=0.00545 moles Ca3(PO4)2

Amount in g = number of moles * molar mass

= 0.00545 moles Ca3(PO4)2* 310.18 g/mol

= 1.69 g Ca3(PO4)2