Calculations involving a limiting reactant
Now consider a situation in which 20.0 g of P4 is added to 54.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations:
1) Calculate the number of moles of PCl5 that can be produced from 20.0 g of P4 (and excess Cl2).
2) Calculate the number of moles of PCl5 that can be produced from 54.0 g of Cl2 (and excess P4).
Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant.
To identify the limiting reactant in a chemical reaction, you need to compare the amounts of the reactants and calculate the moles of the product that can be produced from each reactant. The reactant that produces a lesser amount of product is the limiting reactant. Let's go through the steps to solve this problem:
Step 1: Calculate the moles of PCl5 that can be produced from 20.0 g of P4 (and excess Cl2).
To perform this calculation, we need to know the molar mass of P4. The molar mass of P4 is 123.88 g/mol.
The molar mass conversion factor is:
1 mole P4 = 123.88 g P4
First, calculate the moles of P4:
moles of P4 = mass of P4 / molar mass of P4
moles of P4 = 20.0 g P4 / 123.88 g/mol P4
Step 2: Calculate the moles of PCl5 that can be produced from 20.0 g of P4 (and excess Cl2).
The balanced chemical equation for the reaction is:
P4 + 10Cl2 -> 4PCl5
From the balanced equation, we can see that for every mole of P4, 4 moles of PCl5 are produced. Therefore, the molar ratio is:
1 mole P4 : 4 moles PCl5
Now we can determine the moles of PCl5 that can be produced from 20.0 g of P4:
moles of PCl5 = moles of P4 × molar ratio of PCl5 to P4
moles of PCl5 = moles of P4 × (4 moles PCl5 / 1 mole P4)
Step 3: Calculate the moles of PCl5 that can be produced from 54.0 g of Cl2 (and excess P4).
To perform this calculation, we need to know the molar mass of Cl2. The molar mass of Cl2 is 70.90 g/mol.
The molar mass conversion factor is:
1 mole Cl2 = 70.90 g Cl2
Calculate the moles of Cl2:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
moles of Cl2 = 54.0 g Cl2 / 70.90 g/mol Cl2
Step 4: Calculate the moles of PCl5 that can be produced from 54.0 g of Cl2 (and excess P4).
From the balanced equation, we know that for every 10 moles of Cl2, 4 moles of PCl5 are produced. Therefore, the molar ratio is:
10 moles Cl2 : 4 moles PCl5
Now we can determine the moles of PCl5 that can be produced from 54.0 g of Cl2:
moles of PCl5 = moles of Cl2 × molar ratio of PCl5 to Cl2
moles of PCl5 = moles of Cl2 × (4 moles PCl5 / 10 moles Cl2)
Step 5: Compare the two values of moles of PCl5 calculated in steps 2 and 4. The reactant that produces a smaller amount of PCl5 is the limiting reactant.