How To Balance Cu(No3)2 + Naoh = Cu(Oh)2 + Nano3

I keep rereading what"s in my textbook over & over again, but I"m just lost! Could someone please help & explain! Many thanks!

You"re dealing with a double replacement reaction in which two soluble ionic compounds react to size an insoluble solid that precipitates out of the aqueous solution.

In your case, sodium hydroxide, #"NaOH"#, & copper(II) nitrate, #"Cu"("NO"_3)_2#, will dissociate completely in aqueous solution to form cations & anions

#"NaOH"_text((aq>) -> "Na"_text((aq>)^(+) + "OH"_text((aq>)^(-)#

#"Cu"("NO"_3)_text(2(aq>) -> "Cu"_text((aq>)^(2+) + 2"NO"_text(3(aq>)^(-)#

The reaction will produce copper(II) hydroxide, #"Cu"("OH")_2#, an insoluble ionic compound that precipitates out of solution, & aqueous sodium nitrate, #"NaNO"_3#, another soluble ionic compound.

The balanced chemical equation for this reaction would look lượt thích this

#2"NaOH"_text((aq>) + "Cu"("NO"_3)_text(2(aq>) -> "Cu"("OH")_text(2(s>) darr + 2"NaNO"_text(3(aq>)#

Now, notice that you need #2# moles of sodium hydroxide for every #1# mole of copper(II) nitrate that takes part in the reaction.

Xem thêm: Khai Căn Bậc 2 Của 3 Bằng Bao Nhiêu, Căn Bậc 2 Của 3 Bằng Bao Nhiêu


Xem thêm: CảM NhậN ĐoạN Thơ: “ Tôi Muốn Tắt Nắng Đi, Nghe Thơ: Vội Vàng (Xuân Diệu)

To get the complete ionic equation, rewrite the soluble ionic compounds as cations & anions

#2 xx overbrace(("Na"_text((aq>)^(+) + "OH"_text((aq>)^(-)))^(color(purple)("NaOH")) + overbrace(("Cu"_text((aq>)^(2+) + 2"NO"_text(3(aq>)^(-)))^(color(red)("Cu"("NO"_3)_2)) -> "Cu"("OH")_text(2(s>)# #darr# #+ 2 xx underbrace(("Na"_text((aq>)^(+) + "NO"_text(3(aq>)^(-)))_color(blue)("NaNO"_3)#

This is equivalent to lớn

#2"Na"_text((aq>)^(+) + 2"OH"_text((aq>)^(-) + "Cu"_text((aq>)^(2+) + 2"NO"_text(3(aq>)^(-) -> "Cu"("OH")_text(2(s>) darr + 2"Na"_text((aq>)^(+) + 2"NO"_text(3(aq>)^(-)#

Now, in order to lớn get the net ionic equation, you must eliminate spectator ions, i.e. Ions that are present on both sides of the equation

In this case, you would have

#color(red)(cancel(color(black)(2"Na"_text((aq>)^(+)))) + 2"OH"_text((aq>)^(-) + "Cu"_text((aq>)^(2+) + color(red)(cancel(color(black)(2"NO"_text(3(aq>)^(-)))) -> "Cu"("OH")_text(2(s>) darr + color(red)(cancel(color(black)(2"Na"_text((aq>)^(+)))) + color(red)(cancel(color(black)(2"NO"_text(3(aq>)^(-))))#

which is equivalent khổng lồ

#2"OH"_text((aq>)^(-) + "Cu"_text((aq>)^(2+) -> "Cu"("OH")_text(2(s>) darr#

Copper(II) hydroxide is a blue insoluble solid that precipitates out of solution.