How do you balance #(NH_4)_3PO_4 + Pb(NO_3)_4 - Pb_3(PO_4)_4 + NH_4NO_3#?

Question

How do you balance #(NH_4)_3PO_4 + Pb(NO_3)_4 -> Pb_3(PO_4)_4 + NH_4NO_3#?

Henry Antrim · Accepted Answer

undefined The balanced chemical equation for the reaction is:

(NH₄)₃PO₄ + 4Pb(NO₃)₄ → Pb₃(PO₄)₄ + 12NH₄NO₃

Luna Chen · Answer

#4("NH"_4)_3"PO"_4 +"3Pb"("NO"_3)_4 → "Pb"_3("PO"_4)_4 + "12NH"_4"NO"_3#

The unbalanced equation is

#("NH"_4)_3"PO"_4 +"Pb"("NO"_3)_4 → "Pb"_3("PO"_4)_4 + "NH"_4"NO"_3#

One way to balance the equation is to recognize that the polyatomic ions stay together.

Then we can make some substitutions.

Let #"A" = "NH"_4#; #"X" = "PO"_4#; #"Y" = "NO"_3#.

Then the equation becomes

#"A"_3"X" + "PbY"_4 → "Pb"_3"X"_4 + "AY"#

We start with the most complicated formula, #"Pb"_3"X"_4#, and put a 1 in front of it.

#"A"_3"X" + "PbY"_4 → color(red)(1)"Pb"_3"X"_4 + "AYX"#

Balance #"Pb"#

We have fixed 3 atoms of #"Pb"# on the right, so we need 3 atoms of #"Pb"# on the left. Put a 3 in front of #"PbY"_4#.

#"A"_3"X" + color(blue)(3)"PbY"_4 → color(red)(1)"Pb"_3"X"_4 + "AY"#

Balance #"X"#

We have fixed 4 atoms of #"X"# on the right, so we need 4 atoms of #"X"# on the left. Put a 4 in front of #"A"_3"X"#.

Balance #"A"#

We have fixed 12 atoms of #"A"# on the left, so we need 12 atoms of #"A"# on the right.

#color(orange)(4)"A"_3"X" + color(blue)(3)"PbY"_4 → color(red)(1)"Pb"_3"X"_4 + color(teal)(12)"AY"#

Every formula has a coefficient, and the equation is balanced.

Now, we replace the original formulas in the equation.

The balanced equation is

#4("NH"_4)_3"PO"_4 +"3Pb"("NO"_3)_4 → "Pb"_3("PO"_4)_4 + "12NH"_4"NO"_3#

How do you balance #(NH_4)_3PO_4 + Pb(NO_3)_4 -&gt; Pb_3(PO_4)_4 + NH_4NO_3#?