Question #37ef7

1 Answer
Jan 7, 2015

Answer:

Here's what I got.

Explanation:

First, your general chemical reaction looks like this:

#CuSO_(4(aq)) + Na_2CO_(3(aq)) -> CuCO_(3(s)) darr + Na_2SO_(4(aq))#

Copper(II) sulfate, #CuSO_4#, and sodium carbonate, #Na_2CO_3#, are soluble in aqueous solution, which means that they will dissociate into cations and anions.

The reaction produces copper(II) carbonate, #CuCO_3#, an insoluble solid that precipitates out of solution, and aqueous sodium sulfate, #Na_2SO_4#.

The complete ionic equation is:

#Cu_((aq))^(2+) + SO_(4(aq))^(2-) + 2Na_((aq))^(+) + CO_(3(aq))^(2-) -> CuCO_(3(s)) darr + 2Na_((aq))^(+) + SO_(4(aq))^(2-)#

The net ionic equation, which you get after eliminating spectator ions, is:

#Cu_((aq))^(2+) + color(red)(cancel(color(black)(SO_(4(aq))^(2-)))) + color(red)(cancel(color(black)(2Na_((aq))^(+)))) + CO_(3(aq))^(2-) -> CuCO_(3(s)) darr + color(red)(cancel(color(black)(2Na_((aq))^(+)))) + color(red)(cancel(color(black)(SO_(4(aq))^(2-))))#

#Cu_((aq))^(2+) + CO_(3(aq))^(2-) -> CuCO_(3(s)) darr#