Question

What mass of P4 is needed to produce 85.0 g of PF3 if the reaction has a 64.9% yield? Consider the following reaction: P4 (s) + F2 (g) ----> PF3

Answer 1

To produce 85.0 g of PF3 with a 64.9% yield, you would need 84.89 grams of P4.

Step-by-step explanation:

To determine the mass of P4 needed to produce 85.0 g of PF3 with a 64.9% yield, we need to calculate the theoretical yield of PF3 and then convert it to the mass of P4. The balanced equation for the reaction is:

P4 (s) + F2 (g) → PF3

From the balanced equation, we can see that 1 mol of P4 reacts to produce 4 mol of PF3. First, calculate the molar mass of PF3:

P + 4F = 31.00 g/mol

The molar mass of P4 is 4(31.00 g/mol) = 124.00 g/mol. Now we can use the molar mass and the given mass of PF3 to find the moles of PF3:

moles of PF3 = mass of PF3 / molar mass of PF3

moles of PF3 = 85.0 g / 31.00 g/mol = 2.74 mol

Since the molar ratio between P4 and PF3 is 1:4, the moles of P4 will be:

moles of P4 = 2.74 mol / 4 = 0.685 mol

Finally, we can calculate the mass of P4:

mass of P4 = moles of P4 × molar mass of P4

mass of P4 = 0.685 mol × 124.00 g/mol = 84.89 g.