Question

When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsius, sulfur tetrafluoride is produced. If 25.0g of SF4 were actually produced what is the percent yield

Answer 1

Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

Moles S₈: 256.52g/mol

10.0g * (1mol / 256.52g) = 0.0390 moles

Moles F₂:

PV = nRT

PV/RT = n

Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%