The first thing to perform here is to predict the assets and also compose an unbalanced chemical equation that defines this double replacement reaction.

You recognize that copper(II) sulfate reacts with sodium phosphate, so you deserve to say that

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

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

#"CuSO"_ (4(aq)) -> "Cu"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-)#

#"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)#

At this allude, you should be acquainted with the solubility rules for ionic compounds.

The copper(II) cations will certainly incorporate via the phosphate anions to create 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.

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The second product is aqueous sodium sulfate, #"Na"_ 2"SO"_4#.

The unbalanced chemical equation that explains this reaction looks favor this

#"CuSO"_ (4(aq)) + "Na"_ 3"PO"_ (4(aq)) -> "Cu"_ 3("PO"_ 4)_ (2(s)) darr + "Na"_ 2"SO"_ (4(aq))#

The dvery own arrowhead included alongside copper(II) phosphate indicates that this compound precipitates out of solution.

In order to balance this chemical equation, you should add a coefficient of #color(blue)(3)# to copper(II) sulfate and also sodium sulfate a coreliable of #color(purple)(2)# to sodium phosphate.

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#color(blue)(3)"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(aq))#

Now, you know that #"0.659 g"# of sodium phosphate reacts with excess copper(II) sulfate. This indicates that all the moles of sodium phosphate current in your sample will certainly take part in the reactivity.

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 discover the mass of the precipitate developed by the reaction, usage the reality that you have a #color(purple)(2):1# mole ratio between sodium phosphate and copper(II) phosphate.

The reactivity will certainly develop

#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 transform this to grams, usage the molar mass of copper(II) phosphate

#0.00201 color(red)(cancel(color(black)("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")))#