Limiting Reagent Problem
Discussion: Limiting reagent problems are solved almost the same as regular stoichiometry problems; actually they are two stoichiometry problems rolled into one. How do you recognize the difference between a regular stoichiometry problem and a limiting reagent problem? Simple. Limiting reagent problems have quantities (grams, moles, volume, etc) given in the problem for at least TWO reactants instead of just one. There are at least two methods to solve these problems. Both are shown below.
Example limiting reagent problem
How many grams of CaF2 can be prepared from 37.0 g Ca(OH)2 and 15.0 g HF? (Use molar mass Ca(OH)2 as 74.0, molar mass HF as 20.0, and molar mass CaF2 as 78.0.)
Method number 1:.
Write and balance the equation for the reaction.
Ca(OH)2 + 2HF → CaF2 + 2H2O
Convert the first reactant [Ca(OH)2] to moles.
# moles = grams/molar mass = 37.0/74.0 = 0.500 mole
Convert the other reactant (HF) to moles.
# moles HF = grams/molar mass = 15.0/20.0 = 0.750 mole.
Using the coefficients in the balanced equation, convert 0.500 mole Ca(OH)2 to moles
of the product, CaF2.
# moles CaF2 = 0.500 mole Ca(OH)2 x [1 mole CaF2 /1 mole Ca(OH)2 ] = 0.500 x
(1/1) = 0.500 mole CaF2.
Using the same procedure, convert 0.750 mole HF to moles CaF2.
# moles CaF2 = 0.750 mole HF x [1 mole CaF2 /2 moles HF] = 0.750 x (1/2) = 0.375
USUALLY, the moles from step 3a and 3b are not the same (as in this case); one of
them must be wrong. The correct answer, in limiting reagent problems, is ALWAYS
the smaller value and the reactant producing that value is the limiting reagent.
Therefore, 0.375 mole CaF2 will be produced and HF is the limiting reagent.
Convert moles from step 3c (0.375 mole CaF2) to grams.
moles x molar mass = grams
0.375 mole CaF2 x (78.0 g CaF2 /mole CaF2) = 29.25 g which rounds to 29.2 g). This
is the theoretical yield.
NOTE to method 1: Study this procedure versus the regular stoichiometry procedure (cf http://www.jiskha.com/science/chemistry/stoichiometry.html) and note that if only one reactant amount is given, steps 2b, 3b, and 3c disappear and one is left with the original stoichiometry solution.
Also, see note for method 2 to handle how to calculate the amount of the excess reagent remaining un-reacted.
Method number 2:
Many prefer to use a different method because it is somewhat less work.
Write and balance the equation as in step 1 above.
Convert 37.0 g Ca(OH)2 to moles as in step 2a above to obtain 0.500 mole Ca(OH)2 .
Convert 15.0 g HF to moles as in step 2b above to obtain 0.750 mole HF.
Using the coefficients in the balanced equation convert moles of 2a [moles Ca(OH)2] to
moles of the other reactant (moles HF and not moles of the product].
0.500 mole Ca(OH)2 x [2 moles HF/1 mole Ca(OH)2] = 0.500 x 2 = 1.00 mole HF
required if all of the Ca(OH2 is used. The amount of HF available is 0.750 mole and
not 1.00; therefore, we have too little HF and that is the limiting reagent. All of the HF
will be used; Ca(OH)2, is in excess and some will remain un-reacted.
We know HF is the limiting reagent but let’s check to make sure. Convert moles HF, if
all of the HF is used, to moles Ca(OH)2.
0.750 mole HF x [1 mole Ca(OH)2/2 moles HF] = 0.750 x (1/2) = 0.375 mole Ca(OH)2.
We have more than that; therefore, some of the Ca(OH)2 will remain un-reacted and the
HF is the limiting reagent.
Using the moles of the limiting reagent (HF) convert to moles of the product.
Moles CaF2 = moles HF x (1 mole CaF2/2 moles HF) = 0.750 x (1/2) = 0.375
Convert moles of the product to grams as in step 4 above.
NOTE to method 2: Problems often ask how much of the reagent in excess remains un-reacted. Follow step 3b to convert the limiting reagent to moles of the reagent in excess that is used. Subtract from the initial amount, and convert the final number of moles to grams. In the above example, step 3c converts 0.750 moles HF to 0.375 mole Ca(OH)2 used. 0.375 mole Ca(OH)2 = 0.375 x 74.0 = 27.75 g used. There were 37.0 g initially and 37.0 – 27.75 = 9.25 g which rounds to 9.2 g Ca(OH)2 not reacted.