# Limiting Reagents

#### Lizzie Bartlet

##### Full Member
10+ Year Member
Can someone give me a simple way to figure out limiting reagent problems? The review book I was using tends to present it as a logic type thing and I guess I'm not very logical since I keep getting the wrong answers. I know there's a mathematical way to do it but I've forgotten how.

#### Chemdude

##### Full Member
10+ Year Member
This is the way I think about "limiting reagents." A lot of people tell me I have a strange way of thinking, so I don't know if it'll work for you. I found this problem, let's use it as an example:

Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3

Mg3N2 is 58.1g and the initial amount of H2O is 20.4g

Step 1:Analyze the "actual" ratio of the reactants.

For every Mg3N2 molecule there are six water molecules. So the mole ratio is 1:6=0.167 (the order of the ratio is important)

Step 2: Convert from grams-->moles

Mg3N2:
molar mass(Mg3N2)= 100.93 g/mol

moles= mass/molar mass= 58.1g/100.93 g/mol= 0.576 mol Mg3N2

Water:
molar mass(H2O) = 18.02 g H2O / mol H2O

moles= mass/molar mass= 20.4g/18.02 g/mol= 1.13 mol H2O

Step 3: Analyze the "present" ratio

0.576 mol Mg3N2/ 1.13 mol H2O =0.509

Step 4: Compare "Present" ratio to "actual" ratio

1. If present ratio> actual ratio, then the water is the limiting reagent.

2. If present ratio<actual ratio, then the water is in exess. So the Mg3N2 is the limiting reagent.

3. If present ratio=actual ratio, the there is no limiting reagent.

In this case, present(0.509)>actual ratio(0.167), so water is the limiting reagent.

Hope that helps!!!

#### Bacchus

##### Full Member
Staff member
Volunteer Staff
10+ Year Member
Also you can perform calculations. Using the above and the amounts of starting material: 0.576mol Mg3N2 and 1.13mol H2O.

Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3

You can use each starting amount to see which produces less product Mg(OH)2.

Using Mg3N2:

0.576moles Mg3N2 X (3moles Mg(OH)2/1mole Mg3N2) = 1.728moles Mg(OH)2

Using H2O:

1.13moles H2O X (3moles Mg(OH)2/6moles H2O) = 0.565moles Mg(OH)2

Since water produces less product, it is the limiting reactant.

In the above calculation the amount you're multiplying by is the mole fraction between a reactant (numerator) vs. product (denominator). You get the coefficients from the coefficients in the chemical equation.

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