Question #92fde

!! LONG ANSWER !!

You can balance both chemical equations by inspection, but the first one will be a little more tricky to get right.

Since the first equation describes a redox reaction, you can also balance it by using the oxidation number method.

So, let's start with the second one. Clacium phosphate reacts with sulfuric acid to produce calcium sulfate and calcium dihydrogen phosphate

#"Ca"""_3("PO"_4)_2 + "H"_2"SO"_4 -> "CaSO"""_4 + "Ca"("H"_2"PO"_4)_2#

Notice that you have 4 hydrogen atoms on the products' side, but only 2 on the reactants' side. Multiply the sulfuric acid by 2 to balance the hydrogen atoms out

#"Ca"""_3("PO"_4)_2 + 2"H"_2"SO"_4 -> "CaSO"""_4 + "Ca"("H"_2"PO"_4)_2#

Now notice that you have 2 sulfate anions, #"SO"_4^(2-)#, on the reactants' side and only 1 on the products side. Multiply th calcium sulfate by 2 to get

#"Ca"""_3("PO"_4)_2 + 2"H"_2"SO"_4 -> 2"CaSO"""_4 + "Ca"("H"_2"PO"_4)_2#

And that's it. The calcium atoms are balanced, since you have 3 on the reactantss side and 3 on the products' side. The phosphorus and oxygen atoms are also balanced, with 2 on boths sides and 16 on both sides, respectively.

Now focus on the first equation. Lead (II) nitrate will undergo decomposition to form lead (II) oxide, nitrogen dioxide, and oxygen gas.

Start by assigning oxidation numbers to all the atoms that take part in the reaction

#stackrel(color(blue)(+2))("Pb") (stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3)_2 -> stackrel(color(blue)(+2))("Pb") stackrel(color(blue)(-2))("O") + stackrel(color(blue)(+4))("N") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(0))("O")_2#

Notice that nitrogen's oxidation state goes from +5 on the reactant's side to +4 on the products' side, which means that nitrogen is being reduced.

On the other hand, oxygen's oxidation state goes from -2 on the reactant's side, to 0 on the reactants' side, which means that oxygen is being oxidized.

The half-reactions will look like this

• reduction half-reaction

#2stackrel(color(blue)(+5))("N")"O"_3^(-) + 2e^(-) -> 2stackrel(color(blue)(+4))("N")"O"_2 + 2"O"^(2-)#

In the reduction half-reaction, two nitrogen atoms will gain a total of 2 electrons.

• oxidation half-reaction

#2"O"^(2-) -> stackrel(color(blue)(0))("O")_2 + 4e^(-)#

In the oxidation half-reaction, 2 oxygen atoms lose a total of 4 electrons.

In any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you multiply the reductionm half-reaction by 2 to get a total number of 4 electrons transferred.

#{(4"NO"_3^(-) + 4e^(-) -> 4"NO"_2 + 4"O"^(2-)), (2"O"^(2-) -> "O"_2 + 4e^(-)) :}#

Add the two half-reactions and cancel out what you can to get

#4"NO"_3^(-) + color(red)(cancel(color(black)(4e^(-)))) + color(red)(cancel(color(black)(2"O"^(2-)))) -> 4"NO"_2 + color(red)(cancel(color(black)(4e^(-)))) + "O"_2 + color(red)(cancel(color(black)(4)))"O"^(2-)#

#4"NO"_3^(-) -> 4"NO"_2 + "O"_2 + 2"O"^(2-)#

Be careful when you write the overall reaction because you get 2 nitrate anions from 1 mole of lead (II) nitrate, so multiply lead (II) nitrate by 2, not by 4!

#2"Pb"("NO"_3)_2 -> 2"PbO" + 4"NO"_2 + "O"_2#