Limiting reagent problems are simple to recognize as the initial amounts of more than one species will be given. For instance, for the reaction:

2H₂(g) + O₂(g) 2H₂O

If the amounts of both H₂ and O₂ are given, then this is a limiting reagent problem; otherwise, this is not a limiting reagent problem. There are 6 steps to a limiting reagent problem:

1. Convert the amounts of all species to moles
2. Divide the number of moles of each reactant species by its stoichiometric coefficient. The smallest number corresponds to the limiting reagent.
3. Determine the moles of all other species that react or form by multiplying the smallest number from step 2 by each species' stoichiometric coefficient.
4. Add/Subtract the moles determined in step 3 to/from the initial moles.
5. Convert moles to amounts
6. Check to make sure that total initial mass = total final mass

Here are some practice problems to do before we explain why this approach works. It is easiest to set up a table to keep track of all the values.

In the reaction Mg₃N₂ + 6H₂O 3Mg(OH)₂ + 2NH₃, if the initial amount of Mg₃N₂ is 58.1g and the initial amount of H₂O is 20.4g, what are the final masses of each specie?

1. Convert the amounts of all species to moles

MM(Mg₃N₂) = 100.93 g Mg₃N₂ / mol Mg₃N₂

mass(Mg₃N₂) = 58.1 g Mg₃N₂ x (1 mol Mg₃N₂ / 100.93 g Mg₃N₂) = 0.576 mol Mg₃N₂

MM(H₂O) = 18.02 g H₂O / mol H₂O

mass(H₂O) =20.4 g H₂O x (1 mol H₂O / 18.02 g H₂O) = 1.13 mol H₂O

Compound	Mg3N2	H2O	Mg(OH)2	NH3
MM	100.93	18.02	58.32	17.03
Initial Mass	58.1 g	20.4 g	0.0 g	0.0 g
Initial Moles	0.576 mol Mg3N2	1.13 mol H2O	0.0 mol Mg(OH)2	0.0 mol NH3
Step 2 Division
Step 3 Amounts
Final Moles
Final Mass

1. Divide the number of moles of each reactant species by its stoichiometric coefficient. The smallest amount is the limiting reagent.

Compound	Mg3N2	H2O	Mg(OH)2	NH3
MM	100.93	18.02	58.32	17.03
Initial Mass	58.1 g	20.4 g	0.0 g	0.0 g
Initial Moles	0.576 mol Mg3N2	1.13 mol H2O	0.0 mol Mg(OH)2	0.0 mol NH3
Step 2 Division	0.576 mol / 1 = 0.576 mol	1.13 mol / 6 = 0.189 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts
Final Moles
Final Mass

1. Determine the amount of all other species that react or form by multiplying the smallest number from step 2 by each species stoichiometric coefficient.

Compound	Mg3N2	H2O	Mg(OH)2	NH3
MM	100.93	18.02	58.32	17.03
Initial Mass	58.1 g	20.4 g	0.0 g	0.0 g
Initial Moles	0.576 mol Mg3N2	1.13 mol H2O	0.0 mol Mg(OH)2	0.0 mol NH3
Step 2 Division	0.576 mol / 1 = 0.576 mol	1.13 mol / 6 = 0.189 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	1 x 0.189 mol = 0.189 mol Mg3N2	6 x 0.189 mol = 1.13 mol H2O	3 x 0.189 mol = 0.566 mol Mg(OH)2	2 x 0.189 mol = 0.377 mol NH3
Final Moles
Final Mass

1. Add/Subtract the amounts determined in step 3 to/from the initial amounts.

Compound	Mg3N2	H2O	Mg(OH)2	NH3
MM	100.93	18.02	58.32	17.03
Initial Mass	58.1 g	20.4 g	0.0 g	0.0 g
Initial Moles	0.576 mol Mg3N2	1.13 mol H2O	0.0 mol Mg(OH)2	0.0 mol NH3
Step 2 Division	0.576 mol / 1 = 0.576 mol	1.13 mol / 6 = 0.189 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	1 x 0.189 mol = 0.189 mol Mg3N2	6 x 0.189 mol = 1.13 mol H2O	3 x 0.189 mol = 0.566 mol Mg(OH)2	2 x 0.189 mol = 0.377 mol NH3
Final Moles	0.576 mol - 0.189 mol = 0.387 mol Mg3N2	1.13 mol - 1.13 mol = 0.0 mol H2O	0.0 mol + 0.566 mol = 0.566 mol Mg(OH)2	0.0 mol + 0.377 mol = 0.377 mol NH3
Final Mass

1. Convert moles to amounts

Compound	Mg3N2	H2O	Mg(OH)2	NH3
MM	100.93	18.02	58.32	17.03
Initial Mass	58.1 g	20.4 g	0.0 g	0.0 g
Initial Moles	0.576 mol Mg3N2	1.13 mol H2O	0.0 mol Mg(OH)2	0.0 mol NH3
Step 2 Division	0.576 mol / 1 = 0.576 mol	1.13 mol / 6 = 0.189 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	1 x 0.189 mol = 0.189 mol Mg3N2	6 x 0.189 mol = 1.13 mol H2O	3 x 0.189 mol = 0.566 mol Mg(OH)2	2 x 0.189 mol = 0.377 mol NH3
Final Moles	0.576 mol - 0.189 mol = 0.387 mol Mg3N2	1.13 mol - 1.13 mol = 0.0 mol H2O	0.0 mol + 0.566 mol = 0.566 mol Mg(OH)2	0.0 mol + 0.377 mol = 0.377 mol NH3
Final Mass	0.387 mol x 100.93 g/mol = 39.1 g Mg3N2	0.0 g	0.566 mol x 58.32 g/mol = 33.0 g Mg(OH)2	0.377 mol x 17.03 g/mol = 6.42 g NH3

1. Check to make sure that total initial mass = total final mass
2. + 20.4 = 78.5; 39.1 + 33.0 + 6.42 = 78.5…Mass is conserved!

In the reaction Au₂S₃ + 3H₂ 3H₂S + 2Au, if the initial amount of Au₂S₃ is 500.20 g and the initial amount of H₂ is 5.67g, what are the final masses of each specie?

1. Convert the amounts of all species to moles

MM(Au₂S₃) =490.15 g Au₂S₃ / mol Au₂S₃

mass(Au₂S₃) =500.20 g Au₂S₃x (1 mol Au₂S₃/ 490.15 g Au₂S₃) = 0.1.0204 mol Au₂S₃

MM(H₂) = 2.02 g H₂/ mol H₂

mass(H₂) =5.67 g H₂x (1 mol H₂/ 2.02 g H₂) = 2.8125 mol H₂

Compound	Au2S3	H2	H2S	Au
MM	490.15	2.02	34.09	196.97
Initial Mass	500.20 g	5.67 g	0.0 g	0.0 g
Initial Moles	1.0204 mol Au2S3	2.8125 mol H2	0.0 mol H2S	0.0 mol Au
Step 2 Division
Step 3 Amounts
Final Moles
Final Mass

1. Divide the number of moles of each reactant species by its stoichiometric coefficient. The smallest amount is the limiting reagent.

Compound	Au2S3	H2	H2S	Au
MM	490.15	2.02	34.09	196.97
Initial Mass	500.20 g	5.67 g	0.0 g	0.0 g
Initial Moles	1.0204 mol Au2S3	2.8125 mol H2	0.0 mol H2S	0.0 mol Au
Step 2 Division	1.0204 mol / 1 = 1.0204 mol	2.8125 mol / 3 = 0.9375 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts
Final Moles
Final Mass

1. Determine the amount of all other species that react or form by multiplying the smallest number from step 2 by each species stoichiometric coefficient.

Compound	Au2S3	H2	H2S	Au
MM	490.15	2.02	34.09	196.97
Initial Mass	500.20 g	5.67 g	0.0 g	0.0 g
Initial Moles	1.0204 mol Au2S3	2.8125 mol H2	0.0 mol H2S	0.0 mol Au
Step 2 Division	1.0204 mol / 1 = 1.0204 mol	2.8125 mol / 6 = 0.9375 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	1 x 0.9375 mol = 0.9375 mol Au2S3	3 x 0.9375 mol = 2.8125 mol H2	3 x 0.9375 mol = 2.8125 mol H2S	2 x 0.9375 mol = 1.875 mol Au
Final Moles
Final Mass

1. Add/Subtract the amounts determined in step 3 to/from the initial amounts.

Compound	Au2S3	H2	H2S	Au
MM	490.15	2.02	34.09	196.97
Initial Mass	500.20 g	5.67 g	0.0 g	0.0 g
Initial Moles	1.0204 mol Au2S3	2.8125 mol H2	0.0 mol H2S	0.0 mol Au
Step 2 Division	1.0204 mol / 1 = 1.0204 mol	2.8125 mol / 6 = 0.9375 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	1 x 0.9375 mol = 0.9375 mol Au2S3	6 x 0.9375 mol = 2.8125 mol H2	3 x 0.9375 mol = 2.8125 mol H2S	2 x 0.9375 mol = 1.875 mol Au
Final Moles	1.0204 mol - 0.9375 mol = 0.08288 mol Au2S3	2.8125 mol - 2.8125 mol = 0.0 mol H2	0.0 mol + 2.8125 mol = 2.8125 mol H2S	0.0 mol + 1.875 mol = 1.875 mol Au
Final Mass

1. Convert moles to amounts

Compound	Au2S3	H2	H2S	Au
MM	490.15	2.02	34.09	196.97
Initial Mass	500.20 g	5.67 g	0.0 g	0.0 g
Initial Moles	1.0204 mol Au2S3	2.8125 mol H2	0.0 mol H2S	0.0 mol Au
Step 2 Division	1.0204 mol / 1 = 1.0204 mol	2.8125 mol / 6 = 0.9375 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	1 x 0.9375 mol = 0.9375 mol Au2S3	6 x 0.9375 mol = 2.8125 mol H2	3 x 0.9375 mol = 2.8125 mol H2S	2 x 0.9375 mol = 1.875 mol Au
Final Moles	1.0204 mol - 0.9375 mol = 0.08288 mol Au2S3	2.8125 mol - 2.8125 mol = 0.0 mol H2	0.0 mol + 2.8125 mol = 2.8125 mol H2S	0.0 mol + 1.875 mol = 1.875 mol Au
Final Mass	0.08288 mol x 490.15 g/mol = 40.6 g Au2S3	0.0 g	2.8125 mol x 34.09 g/mol = 95.9 g H2S	1.875 mol x 196.97 g/mol = 369.4 g Au

1. Check to make sure that total initial mass = total final mass

500.20 + 5.67 = 505.87; 40.6 + 95.9 + 369.4 = 505.9…Mass is conserved!

In the reaction 2C₃H₆ + 2NH₃ + 3O₂ 2C₃H₃N + 6H₂O, if the initial amounts are C₃H₆ 22.5g, NH₃ 20.6 g, and O₂ 18.1 g, what are the final amounts of each specie?

1. Convert the amounts of all species to moles

Compound	C3H6	NH3	O2	C3H3N	H2O
MM	42.078	17.034	32.000	53.064	196.97
Initial Mass	22.5 g	20.6 g	18.1 g	0.0 g	0.0 g
Initial Moles	0.5347 mol C3H6	1.2094 mol NH3	0.5656 mol O2	0.0 mol C3H3N	0.0 mol H2O
Step 2 Division
Step 3 Amounts
Final Moles
Final Mass

1. Divide the number of moles of each reactant species by its stoichiometric coefficient. The smallest amount is the limiting reagent.

Compound	C3H6	NH3	O2	C3H3N	H2O
MM	42.078	17.034	32.000	53.064	196.97
Initial Mass	22.5 g	20.6 g	18.1 g	0.0 g	0.0 g
Initial Moles	0.5347 mol C3H6	1.2094 mol NH3	0.5656 mol O2	0.0 mol C3H3N	0.0 mol H2O
Step 2 Division	0.5347 mol / 2 =0.2675 mol	1.2094 mol / 2 = 0.6047 mol	0.5656 mol / 3 = 0.1885 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts
Final Moles
Final Mass

1. Determine the amount of all other species that react or form by multiplying the smallest number from step 2 by each species stoichiometric coefficient.

Compound	C3H6	NH3	O2	C3H3N	H2O
MM	42.078	17.034	32.000	53.064	196.97
Initial Mass	22.5 g	20.6 g	18.1 g	0.0 g	0.0 g
Initial Moles	0.5347 mol C3H6	1.2094 mol NH3	0.5656 mol O2	0.0 mol C3H3N	0.0 mol H2O
Step 2 Division	0.5347 mol / 2 =0.2675 mol	1.2094 mol / 2 = 0.6047 mol	0.5656 mol / 3 = 0.1885 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	2 x 0.1885 mol = 0.3771 mol C3H6	2 x 0.1885 mol = 0.3771 mol NH3	3 x 0.1885 mol = 0.5656 mol O2	2 x 0.1885 mol = 0.3771 mol C3H3N	6 x 0.1885 mol =1.1313 mol H2O
Final Moles
Final Mass

1. Add/Subtract the amounts determined in step 3 to/from the initial amounts.

Compound	C3H6	NH3	O2	C3H3N	H2O
MM	42.078	17.034	32.000	53.064	196.97
Initial Mass	22.5 g	20.6 g	18.1 g	0.0 g	0.0 g
Initial Moles	0.5347 mol C3H6	1.2094 mol NH3	0.5656 mol O2	0.0 mol C3H3N	0.0 mol H2O
Step 2 Division	0.5347 mol / 2 =0.2675 mol	1.2094 mol / 2 = 0.6047 mol	0.5656 mol / 3 = 0.1885 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	2 x 0.1885 mol = 0.3771 mol C3H6	2 x 0.1885 mol = 0.3771 mol NH3	3 x 0.1885 mol = 0.5656 mol O2	2 x 0.1885 mol = 0.3771 mol C3H3N	6 x 0.1885 mol =1.1313 mol H2O
Final Moles	0.5347 - 0.3771 = 0.1576 mol C3H6	1.2094 - 0.3771 = 0.8323 mol NH3	0.5656 - 0.5656 = 0.0 mol O2	0.0 + 0.3771 = 0.3771 mol C3H3N	0.0 + 1.1313 = 1.1313 mol H2O
Final Mass

1. Convert moles to amounts

Compound	C3H6	NH3	O2	C3H3N	H2O
MM	42.078	17.034	32.000	53.064	18.016
Initial Mass	22.5 g	20.6 g	18.1 g	0.0 g	0.0 g
Initial Moles	0.5347 mol C3H6	1.2094 mol NH3	0.5656 mol O2	0.0 mol C3H3N	0.0 mol H2O
Step 2 Division	0.5347 mol / 2 =0.2675 mol	1.2094 mol / 2 = 0.6047 mol	0.5656 mol / 3 = 0.1885 mol… SMALLEST NUMBER SO LIMITING REAGENT
Step 3 Amounts	2 x 0.1885 mol = 0.3771 mol C3H6	2 x 0.1885 mol = 0.3771 mol NH3	3 x 0.1885 mol = 0.5656 mol O2	2 x 0.1885 mol = 0.3771 mol C3H3N	6 x 0.1885 mol =1.1313 mol H2O
Final Moles	0.5347 - 0.3771 = 0.1576 mol C3H6	1.2094 - 0.3771 = 0.8323 mol NH3	0.5656 - 0.5656 = 0.0 mol O2	0.0 + 0.3771 = 0.3771 mol C3H3N	0.0 + 1.1313 = 1.1313 mol H2O
Final Mass	0.1576 mol x (42.078 g /mol) = 6.6 g C3H6	0.8323 mol x (17.034 g/mol) = 14.2 g NH3	0.0 g O2	0.3771 x (53.064 g/mol) = 20.0 g C3H3N	1.1313 x (18.016 g/mol) = 20.4 g H2O

1. Check to make sure that total initial mass = total final mass

22.5 + 20.6 + 18.1 = 61.2; 6.6 + 14.2 + 20.0 +20.4 = 61.2g…Mass is conserved!

Now why do steps 2 and 3 give the correct results?

Give a reaction aA +bB cC + dD, we know that a moles of A will react with b moles of B to give c moles of C and d moles of D. This means, for instance, that Y moles of A will require Y mol A x (b mol B / a mol A) = Yb/a mol B, since the stoichiometric coefficients are conversion factors between different species. We define 1 mol of this reaction as the process of mixing a mol of A with b mol of B to form c mol C and d mol D. Thus, if we have Y mol of A, we can convert to mol of reaction by Y x (1 mol reaction / a mol A) = Y/a. Thus, the divisions in step 2 convert each initial amount into units of mol reaction and, hence, calculate the amount of the reaction that each reactant can permit. The smallest of these amounts is then the least amount of reaction that any initial amount of a reactant will allow. Step 3 is then converting from units of mol reaction to mol of a species by multiplying by Z mol reaction x (c mol C / 1 mol reaction), e.g.