# Question 789cd

Feb 27, 2017

Here's how you can do that.

#### Explanation:

The first thing to do here is to predict the products and write an unbalanced chemical equation that describes this double replacement reaction.

You know that copper(II) sulfate reacts with sodium phosphate, so you can say that

"CuSO"_ (4(aq)) + "Na"_ 3"PO"_ (4(aq)) -> ?

Now, both reactants are soluble ionic compounds, which means that they dissociate completely in aqueous solution to form cations and anions.

${\text{CuSO"_ (4(aq)) -> "Cu"_ ((aq))^(2+) + "SO}}_{4 \left(a q\right)}^{2 -}$

${\text{Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO}}_{4 \left(a q\right)}^{3 -}$

At this point, you should be familiar with the solubility rules for ionic compounds.

The copper(II) cations will combine with the phosphate anions to form copper(II) phosphate, "Cu"_ 3("PO"_4)_2, an insoluble ionic compound that precipitates out of solution.

This is what a precipitate actually is -- an insoluble ionic compound that "falls" out of solution, i.e. it precipitates.

The second product is aqueous sodium sulfate, ${\text{Na"_ 2"SO}}_{4}$.

The unbalanced chemical equation that describes this reaction looks like this

${\text{CuSO"_ (4(aq)) + "Na"_ 3"PO"_ (4(aq)) -> "Cu"_ 3("PO"_ 4)_ (2(s)) darr + "Na"_ 2"SO}}_{4 \left(a q\right)}$

The down arrow added next to copper(II) phosphate indicates that this compound precipitates out of solution.

In order to balance this chemical equation, you must add a coefficient of $\textcolor{b l u e}{3}$ to copper(II) sulfate and sodium sulfate a coefficient of $\textcolor{p u r p \le}{2}$ to sodium phosphate.

$\textcolor{b l u e}{3} {\text{CuSO"_ (4(aq)) + color(purple)(2)"Na"_ 3"PO"_ (4(aq)) -> "Cu"_ 3("PO"_ 4)_ (2(s)) darr + color(blue)(3)"Na"_ 2"SO}}_{4 \left(a q\right)}$

Now, you know that $\text{0.659 g}$ of sodium phosphate reacts with excess copper(II) sulfate. This means that all the moles of sodium phosphate present in your sample will take part in the reaction.

More specifically, the reaction will consume

0.659 color(red)(cancel(color(black)("g"))) * overbrace(("1 mole Na"_ 3"PO"_ 4)/(163.94color(red)(cancel(color(black)("g")))))^(color(blue)("the molar mass of Na"_ 3"PO"_4)) = "0.00402 moles Na"_3"PO"_4

In order to find the mass of the precipitate formed by the reaction, use the fact that you have a $\textcolor{p u r p \le}{2} : 1$ mole ratio between sodium phosphate and copper(II) phosphate.

The reaction will produce

0.00402 color(red)(cancel(color(black)("moles Na"_ 3"PO"_ 4))) * ("1 mole Cu"_ 3("PO"_ 4)_ 2)/(color(purple)(2)color(red)(cancel(color(black)("moles Na"_ 3"PO"_ 4)))) = "0.00201 moles Cu"_ 3("PO"_ 4)_ 2#

To convert this to grams, use the molar mass of copper(II) phosphate

$0.00201 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Cu"_ 3("PO"_ 4)_ 2))) * "380.581 g"/(1color(red)(cancel(color(black)("mole Cu"_ 3("PO"_ 4)_ 2)))) = color(darkgreen)(ul(color(black)("0.765 g}}}}$

The answer is rounded to three sig figs.