Reactions require more than one chemical reagent.
We start with a balanced chemical equation.
Combustion of propane gas:
C3H8 + 6O2 = 6CO2 + 6H2O
In a reaction, the limiting reagent is the reactant, which determines the amount of product produced. The limiting reagent is the chemical, which has the least quantity available before a reaction takes place.
To calculate use stoichiometry.
Grams of A-> Mol A-> Mol B-> Grams of B
We have 3.44g of O2 and 14.8g of C3H8, what is the limiting reagent?
We must consider the moles of the substance to find the limiting reagent. We choose one of the products and find how much of it will be produced if all of each chemical reacts.
3.44g O2 x 1mol O2 x 6 mol H2O x 18g H2O
32g O2 6 mol O2 1 mol H2O
= 1.9 g of H2O produced if all 3.44g of O2 reacts
* If given a value in
M/L instead of just moles , change a ml value into a L and
use dimensional analysis to eliminate
14.8g of C3H8 x 1 mol C3H8 x 6 mol H2O x 18g H2O
44g C3H8 1 mol C3H8 1 mol H2O
= 36.3 g of H2O produced if all 14.8g of C3H8 reacts
O2 is the limiting reagent.
C3H8 is the excess reagent.
The excess reagent is the chemical, which is left over after the reaction.
How many grams of the excess reagent will be left over after the reaction?
1.9g H2O x 1mol H2O x 1 mol C3H8 x 44g C3H8
18g H2O 6 mol H2O 1 mol C3H8
= 0.77g of C3H8
(Started with 14.8g C3H8) – (Used 0.77g of C3H8) = 14.03g of C3H8 left over
This video shows a different method of calculating limiting reagent.
We start with a balanced chemical equation.
Combustion of propane gas:
C3H8 + 6O2 = 6CO2 + 6H2O
In a reaction, the limiting reagent is the reactant, which determines the amount of product produced. The limiting reagent is the chemical, which has the least quantity available before a reaction takes place.
To calculate use stoichiometry.
Grams of A-> Mol A-> Mol B-> Grams of B
We have 3.44g of O2 and 14.8g of C3H8, what is the limiting reagent?
We must consider the moles of the substance to find the limiting reagent. We choose one of the products and find how much of it will be produced if all of each chemical reacts.
3.44g O2 x 1mol O2 x 6 mol H2O x 18g H2O
32g O2 6 mol O2 1 mol H2O
= 1.9 g of H2O produced if all 3.44g of O2 reacts
* If given a value in
M/L instead of just moles , change a ml value into a L and
use dimensional analysis to eliminate
14.8g of C3H8 x 1 mol C3H8 x 6 mol H2O x 18g H2O
44g C3H8 1 mol C3H8 1 mol H2O
= 36.3 g of H2O produced if all 14.8g of C3H8 reacts
O2 is the limiting reagent.
C3H8 is the excess reagent.
The excess reagent is the chemical, which is left over after the reaction.
How many grams of the excess reagent will be left over after the reaction?
1.9g H2O x 1mol H2O x 1 mol C3H8 x 44g C3H8
18g H2O 6 mol H2O 1 mol C3H8
= 0.77g of C3H8
(Started with 14.8g C3H8) – (Used 0.77g of C3H8) = 14.03g of C3H8 left over
This video shows a different method of calculating limiting reagent.