Limiting Reactants method:
Balance your equation first
List all of your information you know in a neat table:
Reaction
CuCl2
2NaNO3
Mass (g)
15
20
1
2
 Cu(NO3)2
2NaCl
ANSWER
Mr (gmol-1)
Moles (mol)
Molar ratio?
1
2
Is this the
limiting reagent?
If anything has an odd ratio (not 1 to 1), then its atoms get made or used up faster than
something without a number on the front.
Next, work out the moles using Mr (molecular mass), which you can calculate from adding
all of the Ar (atomic mass) found in the periodic tables for each individual atom in the
molecule.
Work out which is the limiting reagent (the one with the smallest moles)
Reaction
CuCl2
2NaNO3
Mass (g)
15
20
Mr (gmol-1)
63.5 + 35.5x2 =
86 x 2 (there are
144.5
2 NaNO3 in each
 Cu(NO3)2
2NaCl
ANSWER
reaction)= 172
Moles (mol)
15/144.5 =
20/(86x2) =
0.104
0.116
Molar ratio?
1
2
Is this the
Yes
No
limiting reagent?
1
2
Write the moles that are possible to make, in the moles column of the products.
Using the equation n x Mr = mass, calculate mass, remembering to times through by the
molar ratio.
Reaction
CuCl2
2NaNO3
 Cu(NO3)2
2NaCl (Desired
Product)
Mass (g)
15
20
0.208 x 59.5 = 12.4
(ANSWER)
Mr (gmol-1)
63.5 + 35.5x2 =
24 + 14 + 3x16 =
144.5
86 x 2 (there are
24 + 35.5 = 59.5
2 NaNO3 in each
reaction)= 172
Moles (mol)
15/144.5 =
20/(172) = 0.116
Use the limiting
0.104
reagent and the
molar ratio to find the
moles of product
0.104 x2 = 0.208
Molar ratio?
1
2
Is this the
Yes
No
limiting reagent?
1
2