# Question 209d0

Nov 27, 2014

Start with a balanced equation, which will give you the mole ratios for the reactants and products.

"2Na"_3"PO"_4("aq") + "3Ca(OH")_2("aq") $\rightarrow$ "Ca"_3("PO"_4)_2("s") + "6NaOH"("aq")

Assumption: ${\text{Ca(OH)}}_{2}$ is in excess so that ${\text{Na"_3"PO}}_{4}$ is the limiting reactant.

First, convert the mass of sodium phosphate to moles. The molar mass of ${\text{Na"_3"PO}}_{4}$ is $\text{163.94g/mol}$.

${\text{5.640g Na"_3"PO}}_{4}$ x $\text{1 mol"/"163.94g}$ = ${\text{0.0344028mol Na"_3"PO}}_{4}$

Second, calculate the moles of "Ca"_3("PO"_4)_2 produced by multiplying the moles of ${\text{Na"_3"PO}}_{4}$ times the mole ratio of ${\text{Na"_3"PO}}_{4}$ to "Ca"_3("PO"_4)_2 from the balanced equation, in which "Ca"_3("PO"_4)_2 is on top.

${\text{0.0344028mol Na"_3"PO}}_{4}$ x ${\text{1 mol Ca"_3("PO"_4)_2"/2 mol Na"_3"PO}}_{4}$ = "0.172014mol Ca"_3("PO"_4)_2

Third, convert moles of "Ca"_3("PO"_4)_2 to mass . The molar mass of "Ca"_3("PO"_4)_2 is $\text{310.18g/mol}$.

"0.172014mol Ca"_3("PO"_4)_2 x $\text{310.18g"/"1 mol}$ = "53.36g Ca"_3("PO"_4)_2 (rounded to four significant figures due to four significant figures in $\text{5.640g}$)

Assuming that ${\text{Ca(OH)}}_{2}$ was in excess, ${\text{5.640g Na"_3"PO}}_{4}$ will produce "53.36g Ca"_3("PO"_4)_2# in the reaction described above.