# Percent yield problem of Chemistry

When 20.0 L of fluorine gas at STP is slowly bubbled throught 100.0g of liquid boron triiodide, 7.55 g of boron trifluoride is produced. What is the percent yield?

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First of all, let's see the equation for the reaction:

3F2 + 2BI3 → 2BF3 + 3I2

From the equation, we can see that 3 moles of fluorine gas react with 2 moles of boron triiodied to give 2 moles of boron trifluoride.

Then, number of moles of fluorine gas present = 20/22.4 = 0.893 moles

Number of moles of boron triiodide present = 100/(10.8 + 3 × 127) = 0.255 moles

Therefore, we can see that boron triiodide is now the limiting reactant in which 0.255 × (3/2) = 0.383 moles of fluorine gas has been reacted.

Hence, from the above deductions, we can see that 0.255 moles of boron trifluoride can be produced theoretically.

Now, number of moles of boron trifluoride produced = 7.55/(10.8 + 3 × 19) = 0.111 moles.

Therefore the percentage yield of boron trifluoride is:

0.111/0.255 × 100% = 43.6%

資料來源： My chemical knowledge