No3 And Ionic Reaction Lab

Balance the following chemical equations and separate the coefficients by commas and a space respectively (put a 1 if the coefficient is assumed 1):

Mg(s)+P4(s)→Mg3P2(s)

Explanation:

Al(s)+N2(s)→AlN(s)

Explanation:

5.2.1 Balancing with polyatomic ions
We can balance each atom like we did in the previous example but sometimes it becomes very cumbersome especially when we have the same atom in different compounds. Let’s look at the following reaction:

YOU CAN DO THIS, BUT IT WILL BE TEDIOUS:

Na2SO4(aq) + Mg(NO3)2(aq) → MgSO4(s) + NaNO3(aq)
2 Na atoms

1 S atom

1 Mg atom

2 N atoms

10 O atoms (4 from SO4 and 6 from (NO3)2)

1 Na atom

1 S atom

1 Mg atom

1 N atom

7 O atoms (4 from SO4 and 3 from NO3)

This will become

very difficult if we balance by oxygen but since SO42- and NO3- remain intact on both sides of the reaction, we can balance for them instead of each atom. This will also alleviate the need to balance for N and S but we will still need to balance for Mg and Na since they are not part of a polyatomic ion.

SO DO THIS INSTEAD:

Na2SO4(aq) + Mg(NO3)2(aq) → MgSO4(s) + NaNO3(aq)

2 Na atoms

1 SO42– ion

1 Mg atom

2 NO3– ions

1 Na atom

1 SO42– ion

1 Mg atom

1 NO3– ion

Since SO4 and Mg are balanced we will start with Na. You could start with NO3 if you want, I just chose Na.

Na2SO4(aq) + Mg(NO3)2(aq) → MgSO4(s) + 2 NaNO3(aq)

2 Na atoms

1 SO42– ion

1 Mg atom

2 NO3– ions

2 Na atoms

1 SO42– ion

1 Mg atom

2 NO3– ions

By balancing Na, we balanced NO3 too and our equation is balanced.

This will often be the case with ionic reactions but not always. Remember there is no magic trick that works all the time.

Let’s take a look at this reaction:

(NH4)3PO4(aq) + Ca(ClO4)2(aq) → Ca3(PO4)2(s) + NH4ClO4(aq)

3 NH4+ ions

1 PO43– ion

1 Ca atom

2 ClO4– ions

1 NH4+ ion

2 PO43– ions

3 Ca atoms

1 ClO4– ion

Since NH4+ and ClO4– are in a one to one ratio for NH4ClO4 on the products side and there are 3 NH4+ ions and 2 ClO4– ions on the reactants side we can put a 6 in front of NH4ClO4 and then balance to make 6 NH4+ ions and 6 ClO4– ions by putting a 2 in front of (NH4­)3PO4 and a 3 in front of Ca(ClO4)2 and see how that works.

2 (NH4)3PO4(aq) + 3 Ca(ClO4)2(aq) → Ca3(PO4)2(s) + 6 NH4ClO4(aq)

6 NH4+ ions

2 PO43– ions

3 Ca atoms

6 ClO4– ions

6 NH4+ ions

2 PO43– ions

3 Ca atoms

6 ClO4– ions

By balancing the two polyatomic ions the rest of the ions are balanced as well. Again, this is often the case but not always.

Quick Check: 5.2.1.a
Balance the following chemical equations and separate the coefficients by commas and a space respectively (put a 1 if the coefficient is assumed 1):

Li2CO3(aq)+Fe2(SO4)3(aq)→Fe2(CO3)3(s)+Li2SO4(aq)

Explanation:

AlCl3(aq)+K3PO4(aq)→AlPO4(s)+KCl(aq)

Explanation:

Balancing reactions with covalent compounds can be a little tricky because sometimes an atom will appear in more than one compound either on the product or reactants side.

Hydrazine (N2H4) is a flammable liquid used in rocket fuel. Balance the chemical reaction which shows its formation from nitrogen gas and ammonia.

NH3(l) + N2(g) → N2H4(l)

3 N atoms

3 H

atoms

2 N atoms

4 H atoms

Since N is in two molecules on the reactants side we will balance for H first. In general, when an atom is present in two or more compounds or molecules, you balance it last. We see that hydrogen comes in groups of 3 and 4 in NH3 and N2H4, respectively, so we will place a 4 in front of NH3 and a 3 in front of N2H4.

4 NH3(l)
+

N2(g)
→

3 N2H4(l)

6 N atoms

12 H atoms

6 N atoms

12 H atoms

By doing that, N became balanced as well.

Propane (C3H8) is used for cooking on a grill outside. It undergoes what is called a combustion reaction to form CO2 and H2O. Balance the chemical reaction:

C3H8(g)
+

O2(g) →
CO2(g)
+

H2O(l)

3 C atoms

8 H atoms

2 O atoms

1 C atom

2 H atoms

3 O atoms

Like before since we see that O is in two different places on the products side, we will balance that last. Also since there is only one C in CO2 and 3 Cs and 8 Hs in C3H8 and 2 Hs in H2O we will balance C first by putting a 3 in front of CO2.

C3H8(g)
+

O2(g)
→

3 CO2(g)
+

H2O(l)

3 C atoms

8 H atoms

2 O atoms

3 C atoms

2 H atoms

7 O atoms

Now let’s balance H. We see there are 8 Hs on the reactants side and 2 Hs on the products side so we will put a 4 in front of water.

C3H8(g) + O2(g) → 3 CO2(g) + 4 H2O(l)

3 C atoms

8 H atoms

2 O atoms

3 C atoms

8 H atoms

10 O atoms

Now we have 10 O atoms on the products side so we will put a 5 in front of the O2 to give us a total of 10 O atoms.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

3 C atoms

8 H atoms

10 O atoms

3 C atoms

8 H atoms

10 O atoms

Carbon monoxide and nitrogen monoxide are toxic gases and are by-products of fuel combustion. The catalytic converter in your car converts them into carbon dioxide and nitrogen which are not toxic. Balance the chemical reaction:

NO(g) + CO(g) → N2(g) + CO2(g)

1 N atom

2 O atoms

1 C atom

2 N atoms

2 O atoms

1 C atom

Since the only atom not balanced is N we will start with that but we can see that since oxygen is in two of the reagents anything we do to N and C will affect O, so we will worry about O last.

2 NO(g) + CO(g) → N2(g) + CO2(g)

2 N atoms

3 O atoms

1 C atom

2 N atoms

2 O atoms

1 C atom

We now have N and C balanced but not O. We need one more O on the product side but can only increase the number of O atoms on the products side in multiple of 2s. So if we put a 2 in front of CO2 and a 2 in front of CO we will have all our atoms balanced.

2 NO(g) + 2 CO(g) → N2(g) + 2 CO2(g)

2 N atoms

4 O atoms

2 C atoms

2 N atoms

4 O atoms

2 C atoms

Quick Check: 5.2.1.b