Untitled - Kelly Walsh High School

Mass, Moles, and Equations 43
the balanced chemical equation. (This is one place where, if we did not balance
the equation, we would be in trouble.)
I 2 0.295508 mol I 2
1
F 2 0.66423 mol F 2
5
The quantity (mole-to-coefficient ratio) of F 2 is smaller than that of I 2; therefore,
fluorine in the limiting reactant (reagent). Once we know the limiting
reactant, all remaining calculations will depend on the limiting reactant.
Now we can find the maximum number of moles of IF 5 that can be produced
(theoretical yield). We will begin with our limiting reactant rather than using
the 45.2 g of IF 5 given (actual yield). Remember, the limiting reagent is our key,
and we don’t want to lose our key once we have found it.
To find the moles of IF 5 from the limiting reagent, we need to use a mole ratio
derived from information in the balanced chemical equation. (This is another
place where, if we had not balanced the equation, we would be in trouble.)
We will not actually calculate the moles of IF 5 yet; we will simply save this setup.
The problem asks for the percent yield. Recall the definition of percent yield:
%yield
0.295508 (unrounded)
0.132846 (unrounded) (limiting reactant)
mol IF5 (0.66423 mol F2 )a 2 mol IF5 b
5 mol F2 actual yield
theoretical yield
In this problem, the actual yield is the amount of product found by the scientist
(45.2 g IF 5); therefore, we need the theoretical yield to finish the problem.
Since our actual yield has the units “g IF 5,” our theoretical yield must have
identical units. To determine the grams of IF 5 from the moles of limiting reactant,
we need the molar mass of IF 5 [126.9 g I/mol I 5 (19.00 g F/mol F)
221.9 g IF 5/mol].
Mass IF 5 (0.66423 mol F 2 )a 2 mol IF 5
5 mol F 2
100 %
58.957 g IF 5 (unrounded)
ba 221.9 g IF5 b
1 mol IF5