Answer : The order of steps necessary to calculate the mass of product formed in a chemical reaction given the masses of all reactants :
(1) Determine the number of moles of each reactant by dividing its mass by its molar mass.
(2) Identify the limiting reactant as the reactant that produces the least amount of product.
(3) Determine the amount of product that could be formed from the limiting reactant using appropriate mole ratio.
(4) Determine the mass, multiply the number of moles of product formed (from the limiting reactant) by the molar mass of the product.
For example :
If we are given that 10 grams of magnesium react with 6 grams of oxygen to give magnesium oxide then calculate the mass of product formed in a chemical reaction.
Solution :
Mass of Mg = 10 g
Mass of
= 6 g
Molar mass of Mg = 24 g/mole
Molar mass of
= 32 g/mole
Molar mass of MgO = 40 g/mole
First we have to calculate the moles of Mg and
.


Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 1 mole of
react with 2 mole of

So, 0.1875 moles of
react with
moles of

From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of

From the reaction, we conclude that
As, 1 mole of
react to give 2 mole of

So, 0.1875 moles of
react to give
moles of

Now we have to calculate the mass of



Hence, the mass of product is 15 grams.