How do you solve limiting reagent problems?

Answer 1

AnswerUnderstanding limiting reagent problems, and being able to solve them, is essential for determining how much of each reactant is needed when performing a reaction, and will also tell you how much of each product will be formed in the reaction. The amount of products formed is determined by the limiting reagent.

Before solving a limiting reagent problem, you MUST first do a couple of things: you must write the balanced chemical reaction and you must determine the stoichiometry of the reaction. Instructions on how to do both of these things can be found under the Related Questions links to the left.

Once you have written the balanced the reaction, and determined the stoichiometry, you must compare two (or more) calculations to determine which reagent is the limiting reagent and which one is in excess.

The limiting reagent is the chemical which will effectively determine the amount of products that are formed. In other words, the limiting reagent is the chemical which will run out first as the reaction occurs and the reactants are consumed. The other reactants, the ones are are leftover when the limiting reagent runs out, are said to be in excess.

Which chemical is limiting and which is in excess is ALWAYS a function of the stoichiometry and the number of moles, and NEVER the number of grams of the reagents. You must always compare the numbers of moles with the stoichiometry of the reaction.

Let me explain how to solve the problem with an example.

Suppose 2.00g of NaCl reacts with 5.00g of AgNO3 to form NaNO3 and AgCl. First we must write the balanced equation, which is:

NaCl + AgNO3 ---> NaNO3 + AgCl

Now we must determine the stoichiometry, which is this case is very simple. The ratio of all species in the reaction is one-to-one. In other words, for one mole of NaCl, one mole of AgNO3 reacts with it, and one mole of NaNO3 and one mole of AgCl are formed.

Now we must determine which reagent is limiting: the NaCl or the AgNO3. As I said, everything must be in terms of moles, and so we must convert the number of grams of each into moles:

2.00 grams of NaCl ÷ 58.4425 grams/mole = 0.03422 moles NaCl

5.00 grams of AgNO3 ÷ 169.987 grams/mole = 0.02944 moles AgNO3

The stoichiometry of the reaction says that the number of moles of NaCl and AgNO3 should be equal, but as you can see, in this case they are not the same. There are more moles of NaCl than AgNO3. That means that AgNO3 is the limiting reagent, and NaCl is in excess because there is more NaCl than can react with AgNO3. That means that 0.02944 moles of NaCl will react with 0.02994 moles of AgNO3, and 0.02944 moles of NaNO3 and 0.02944 moles of AgCl will be formed.

From this we can say how much NaCl will be left one: the amount that we started with minus the amount that reacted:

0.03422 - 0.02944 = 0.00478 moles of NaCl in excess.

Note that NaCl is in excess even though there were fewer grams of NaCl than there were of AgNO3! It's about MOLES and notgrams.

Let us do another example with more complicated stoichiometry.

Suppose 20.0 grams of magnesium hydroxide, Mg(OH)2, reacts with 40.0 grams of phosphoric acid, H3PO4 to form magnesium phosphate, Mg3(PO4)2, and water, H2O. The balanced reaction is:

3Mg(OH)2 + 2H3PO4 --> Mg3(PO4)2 + 6H2O

As you can see the stoichiometry here is more complicated, and the ratios are 3 to 2 to 1 to 6. Again, the next step is to convert grams to moles.

20.0 grams Mg(OH)2 ÷ 58.324 grams/mole = 0.3429 moles Mg(OH)2

40.0 grams H3PO4 ÷ 97.990 grams/mole = 0.4082 moles H3PO4

As you can see, there is more H3PO4 than Mg(OH)2, so it may seem that it is in excess but always reason out your answer and check it using stoichiometry. The stoichiometry of the problem says that for every 3 moles of Mg(OH)2, there needs to be 2 moles of H3PO4. Therefore the limiting reagent is Mg(OH)2 because there clearly isn't 3/2 (1.5 times) the number of moles of H3PO4 provided. Mg(OH)2 is the limiting reagent and H3PO4 is in excess. To check this (if you can't do the math in your head), you can do the calculation using the 2-to-3 stoichiometry of the reaction:

0.3429 moles Mg(OH)2 * 2 mol H3PO4/3 mol Mg(OH)2 = 0.23 moles H3PO4 required

and clearly 0.4082 is more than 0.23.

You can also check it this way: there should be 3/2 (1.5 times) as much Mg(OH)2 as H3PO4:

0.4082 moles H3PO4 * 3 mol Mg(OH)2/2 mol H3PO4 = 0.32 moles Mg(OH)2 required

and since 0.3429 (moles provided) is less than 0.32 (moles needed) Mg(OH)2 is the limiting reagent and H3PO4 is in excess.

Because Mg(OH)2 is the limiting reagent, the amount of products formed is determined by the number of moles of the limiting reagent in this case the Mg(OH)2. For each mole of Mg(OH)2, 1/3 moles of Mg3(PO4)2 is formed (because of the 3-to-1 ratio), and 2 moles of H2O are formed (because of the 3-to-6 ratio). So to find the number of moles of products, multiply the number of moles of Mg(OH)2 by 1/3 and 2 respectively:

0.3429 moles Mg(OH)2 * 1/3 = 0.1143 moles Mg3(PO4)2 formed

0.3429 moles Mg(OH)2 * 2 = .6858 moles H2O formed

Student 09 12:41, 20 Aug 2009 (UTC)sorry

I think the answer wasnt right.

Suppose to be every 3 moles of Mg(OH)2, there needs to be 2 moles of H3PO4.

finally the answer is Mg(OH)2 will be the limiting reagent.

This is my own opinion and i think this is the correct one. i jope that someone can please check on this. if me wrong,do correct me please,,thank you

not sure what the person who wrote out all the problems did for the second example, but after you find the moles of both reactants, you're supposed to divide the number of moles by the coefficient. my calculations were a bit off from the other person's, but this is pretty much what i got:

20.0g Magnesium hydroxide= 0.34 moles * 3 (the coefficient) = 1.02

40.0g phosphoric acid= 0.41 moles * 2 (the coefficient) = 0.82

1.02>0.82, therefore, the phosphoric acid must be limiting.