Step 3: Look for elements that appear only once on each side of the equation but in unequal
numbers of atoms. Balance these elements.

23 NH 32 " H O

Step 4: Balance elements that appear in two or more formulas on the same side of the equation.

Step 5: Double check your balanced equation and be sure the coefficients are the lowest possi-
ble whole numbers.

23 3NH 322 +++CuO" Cu N 3 H O

2 + 3 + 3 + 1 + 3 = 12

(Be sure to include the unwritten 1 that is in front of N 2 .)

4. When 0.600 mole of BaCl2(aq)is mixed with 0.250 mole of K 3 AsO4(aq), what is the
   maximum number of moles of solid Ba 3 (AsO 4 ) 2 that could be formed?

A. 0.125 mole

B. 0.200 mole

C. 0.250 mole

D. 0.375 mole

E. 0.500 mole

Answer: A

Begin by writing a balanced equation:

32 BaCl 24342 ()aq+ K AsO 5 ()aq"Ba AsO()()s+ 6 KCl()aq

You may want to write the net-ionic equation:

32 Ba^23 +-()aq + AsO (^4) ()aq"Ba AsO 34 ^ h (^2) ]sg
Next, realize that this problem is a limiting-reactant problem. That is, one of the two reactants
will run out first, and when that happens, the reaction will stop. You need to determine which
one of the reactants will run out first. To do this, you need to be able to compare them on a 1:1
basis. But their coefficients are different, so you need to relate both reactants to a common
product, say Ba 3 (AsO 4 ) 2. Set the problem up like this:

. ()
.()

mole BaC

moles BaC

mole Ba AsO mole Ba AsO

1

0 600 1

31

(^2132) 0 200
2
4

= 342

. ()
.()

mole K AsO

moles K AsO

mole Ba AsO mole Ba AsO

1

0 250

2

(^3412) 0 125
34
34

= 342

Part II: Specific Topics