Six solved "balancing by groups" problems

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What balancing by groups means is that you balance polyatomic groups (like sulfate or nitrate) as an entire group, rather than looking at the individual atoms that make up the group.

Note that there are also individual elements (in addition to those in the group) that must be balanced as well.

What happens, say in hydrogen for example, is that some of the hydrogen gets balanced in the group and some gets balanced outside of the group. This happens in example 6 where some of the hydrogen is balance via the ammonium group and some is balanced via a coefficient infront of the nitric and and the water.


Problem #1: Al2(SO4)3 + Ca(OH)2 ---> Al(OH)3 + CaSO4

Solution:

1) Balance the sulfate:

Al2(SO4)3 + Ca(OH)2 ---> Al(OH)3 + 3CaSO4

2) Balance the calcium:

Al2(SO4)3 + 3Ca(OH)2 ---> Al(OH)3 + 3CaSO4

3) Balance the hydroxide:

Al2(SO4)3 + 3Ca(OH)2 ---> 2Al(OH)3 + 3CaSO4

Alternate answer:

1) Balance the hydroxides with two coefficients at the same time:

Al2(SO4)3 + 3Ca(OH)2 ---> 2Al(OH)3 + CaSO4

Because 3 x 2 = 2 x 3.

2) The above also balanced the aluminum, now balance the calcium:

Al2(SO4)3 + 3Ca(OH)2 ---> 2Al(OH)3 + 3CaSO4

This balances the sulfate and it's done.


Problem #2: Na3PO4 · 12H2O + Zn(C2H3O2)2 · 2H2O ---> Zn3(PO4)2 · 4H2O + NaC2H3O2 + H2O

Solution:

1) Balance the zinc:

Na3PO4 · 12H2O + 3Zn(C2H3O2)2 · 2H2O ---> Zn3(PO4)2 · 4H2O + NaC2H3O2 + H2O

2) Balance the acetate:

Na3PO4 · 12H2O + 3Zn(C2H3O2)2 · 2H2O ---> Zn3(PO4)2 · 4H2O + 6NaC2H3O2 + H2O

3) Balance sodium:

2Na3PO4 · 12H2O + 3Zn(C2H3O2)2 · 2H2O ---> Zn3(PO4)2 · 4H2O + 6NaC2H3O2 + H2O

4) The above step also balances the phosphates. Balance the water:

2Na3PO4 · 12H2O + 3Zn(C2H3O2)2 · 2H2O ---> Zn3(PO4)2 · 4H2O + 6NaC2H3O2 + 26H2O

Comment on water's coefficient of 26: 2 x 12 = 24 and 3 x 2 = 6 for 30 H2O on the left. On the right, there are 4 in the zinc phosphate, so 26 needed in front of the H2O


Problem #3: (NH4)3PO4 + Ni(NO3)2 ---> Ni3(PO4)2 + NH4NO3

Solution:

1) Balance the Ni:

(NH4)3PO4 + 3Ni(NO3)2 ---> Ni3(PO4)2 + NH4NO3

2) Balance the nitrate:

(NH4)3PO4 + 3Ni(NO3)2 ---> Ni3(PO4)2 + 6NH4NO3

3) Balance the ammonium:

2(NH4)3PO4 + 3Ni(NO3)2 ---> Ni3(PO4)2 + 6NH4NO3

This last step also balances the phosphates.


Problem #4: (NH4)3PO4 + Pb(NO3)4 ---> Pb3(PO4)4 + NH4NO3

Solution:

This problem is analogous to example #3. Balance the Pb, then the nitrate, with the ammonium last. I decided to not put the full answer down, but the sum of all four coefficients in the balanced equation is 20.


Problem #5: Ca3(PO4)2 + H2SO4 ---> CaSO4 + Ca(H2PO4)2

Solution:

1) See how there are two H2PO4 on the right-hand side? Look at just the PO4 and see that PO4 is balanced. Now, balance the H2 part, like this:

Ca3(PO4)2 + 2H2SO4 ---> CaSO4 + Ca(H2PO4)2

2) Now, you have to balance the sulfate:

Ca3(PO4)2 + 2H2SO4 ---> 2CaSO4 + Ca(H2PO4)2

and in so doing, you balance the calcium and it's done.


Problem #6: H3PO4 + (NH4)2MoO4 + HNO3 ---> (NH4)3PO4 · 12MoO3 + NH4NO3 + H2O

Solution:

1) Start with the MoO4 going to MoO3:

H3PO4 + 12(NH4)2MoO4 + HNO3 ---> (NH4)3PO4 · 12MoO3 + NH4NO3 + H2O

Keep in mind that there are 12 oxygen atoms "out there" that will have to be balanced on the right-hand side.

2) Balance the ammonium:

H3PO4 + 12(NH4)2MoO4 + HNO3 ---> (NH4)3PO4 · 12MoO3 + 21NH4NO3 + H2O

3) Balance the nitrate:

H3PO4 + 12(NH4)2MoO4 + 21HNO3 ---> (NH4)3PO4 · 12MoO3 + 21NH4NO3 + H2O

4) Balance the 24 hydrogens (3 in H3PO4 and 21 in HNO3):

H3PO4 + 12(NH4)2MoO4 + 21HNO3 ---> (NH4)3PO4 · 12MoO3 + 21NH4NO3 + 12H2O

Remember those 12 oxygens that were "out there?" The 12 in front of the H2O balances them and it's done.

Notice I never touched the phosphate. That's because it was balanced at the start and it was never affected by any of the balacing moves I made.


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