Hydrogen Fluoride is produced from the reaction of calicum fluoride are reacted with an excess of sulfuric acid. How much hydrogen fluoride is formed if the percent yield is 93.0%?
You don't have a complete problem.
To determine the amount of hydrogen fluoride formed if the percent yield is 93.0%, we need to know the starting amount of calcium fluoride and the balanced chemical equation for the reaction.
Let's assume we have x moles of calcium fluoride (CaF2).
The balanced equation for the reaction is as follows:
CaF2 + H2SO4 -> CaSO4 + 2HF
According to the stoichiometry of the reaction, we know that for every 1 mole of calcium fluoride (CaF2), 2 moles of hydrogen fluoride (2HF) are formed.
So, if x moles of calcium fluoride react completely, we would expect 2x moles of hydrogen fluoride to be formed.
However, the percent yield is given as 93.0%. Percent yield is the ratio of the actual yield to the theoretical yield, multiplied by 100%. In this case, the actual yield is the amount of hydrogen fluoride produced, and the theoretical yield is the maximum amount of hydrogen fluoride that could be produced based on the stoichiometry of the reaction.
We can use the percent yield to calculate the actual yield of hydrogen fluoride. If the percent yield is 93.0%, it means that only 93.0% of the expected yield is obtained.
Actual yield = percent yield x theoretical yield
Actual yield = 0.93 x 2x
Actual yield = 1.86x
Therefore, the amount of hydrogen fluoride formed if the percent yield is 93.0% is 1.86 times the amount of calcium fluoride used in the reaction.