How many grams of fluorine are needed to produce 120. g of phosphorus trifluoride if the reaction has a 78.1% yield from the following reaction: P4+6F2 -> 4PF3
To determine the amount of grams of fluorine needed to produce 120 g of phosphorus trifluoride, we need to follow several steps.
Step 1: Write down the balanced chemical equation for the reaction:
P4 + 6F2 -> 4PF3
Step 2: Calculate the molar mass of phosphorus trifluoride (PF3):
Phosphorus (P) has a molar mass of 30.97 g/mol
Fluorine (F) has a molar mass of 18.998 g/mol
Since we have 3 fluorine atoms in PF3, we multiply the molar mass of fluorine by 3:
Molar mass of PF3 = (30.97 + (18.998*3)) g/mol
Molar mass of PF3 = 87.965 g/mol
Step 3: Calculate the theoretical yield of PF3:
Theoretical yield is the amount of product that should be obtained under ideal conditions.
In the balanced equation, we see that 1 mole of P4 produces 4 moles of PF3.
This means that to produce 87.965 g of PF3, we need (87.965/4) moles of P4, which is equal to 21.99125 moles of P4.
Step 4: Calculate the actual yield of PF3:
Actual yield is the amount of product obtained in reality.
Given that the reaction has a 78.1% yield, we multiply the theoretical yield by 0.781 to obtain the actual yield:
Actual yield = 21.99125 moles * 0.781
Step 5: Calculate the grams of fluorine needed:
In the balanced equation, we see that 6 moles of F2 react with 1 mole of P4 to produce 4 moles of PF3.
Therefore, to produce 21.99125 moles of PF3, we need (21.99125 * 6) moles of F2, which is equal to 131.9475 moles of F2.
Finally, we convert moles of F2 to grams by multiplying by the molar mass of fluorine:
Grams of fluorine needed = 131.9475 moles * 18.998 g/mol
Calculating the final result:
Grams of fluorine needed = 131.9475 * 18.998
Grams of fluorine needed ≈ 2,508.41 g
Therefore, approximately 2,508.41 grams of fluorine are needed to produce 120 grams of phosphorus trifluoride with a 78.1% yield.
Here is a worked example of a stoichiometry problem. Just follow the steps. http://www.jiskha.com/science/chemistry/stoichiometry.html
You need to start with 120/0.781 = ?g to end up with 120 g at 78.1% yield.