## Chem – Limited Reactant

What is LIMITED REACTANT (REAGENT)?

Many students have tremendous trouble with this section. However, it should not be so if they have studied the previous section of combining stoichiometry and molar mass conversions. The only difference between limited reactant and previous conversion sections are that limited reactant problems have 2 or more starting points. This means that you usually have to do 2 or more conversions. The same conversions as you have done before, using the same conversion map as the last section. Again, we can turn back to the baking analogy for real life example of limited reactant. If we have the same baking equation below as we had before and I tell you that we start with 5 eggs and 7 cups of sugar, how many cookies can we make? Which ingredient runs out first (which reactant is limiting)? Which ingredient will we have extra after the cookies are made (which reactant will be in excess)?

2 eggs + 3 cups of sugar —> 24 cookies

In this case we can make 56 cookies since the 7 cups of sugar will be the first ingredient to run out. That means that we will have some egg left over because we did not use all of them up.

 7 cups of sugar 24 cookies = 56 cookies 3 cups of sugar

So the limited reactant is the chemical that will be used up first in a reaction. The excess reactant is the one that will have some amount remaining after the reaction is done.

VIDEO Stoichiometry (Limited Reactant) Conversions Demonstrated Example 6: If you have 12 g of N2 and 3g of H2 then what is your limited reactant? How many grams of NH3 can you make? You will need a periodic table to help solve this problem.

N2(g) + 3 H2(g) —-> 2 NH3(g)

Step 1:

What information does the question supply us with?

Answer: 12 g of N2 and 3g of H2

Step 2:

What units does the question ask?

Answer: the limited reactant and g NH3

Step 3:

Where do we start?

Answer: With either of the pieces of information the questions supplies us with. I will start with 12g of N2.

Step 4:

How many conversions must we do?

Answer: Look at the conversion map. We pass through 3 arrows when we go from Grams of A —> moles of A —> moles of B —> Grams of B. 3 arrows = 3 conversions

Step 5:

How do we set up the first set of conversions for this problem?

 12 g N2 g NH3 1

Step 6:

What is the first conversion?

Answer: molar mass (grams to mole ratio) of N2 found on the periodic table

Step 7:

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 12 g N2 mol g NH3 g

Step 8:

What is the next step?

Answer: Fill in the numbers and cross out units

 12 g N2 1 mol g NH3 28 g

Step 9:  Simplify

 12 N2 1 mol g NH3 28

Step 10:

What is the next conversion?

Answer: mole to mole ratio (coefficient ratio)

Step 11:

How do I set it up?

Answer: units first, set up the units that you need to cancel out (in red)

 12 N2 1 mol NH3 g NH3 28 N2

Step 12:

What is the next step?

Answer: Fill in the numbers and cross out units

 12 N2 1 mol 2 NH3 g NH3 28 1 N2

Step 13:  Simplify

 12 1 mol 2 NH3 g NH3 28 1

Step 14:

What is the next conversion?

Answer: molar mass (grams to mole ratio) of NH3 found on the periodic table

Step 15:

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 12 1 mol 2 NH3 g = g NH3 28 1 mol

Step 16:

What is the next step?

Answer: Fill in the numbers and cross out units

 12 1 mol 2 NH3 17 g = g NH3 28 1 1 mol

Step 17:  Simplify

 12 1 2 NH3 17 g = g NH3 28 1 1

Step 18:

How do I know I am done with this conversion?

Answer: The only units left are the units that match the answer. In this case g and NH3

 12 1 2 NH3 17 g = g NH3 28 1 1

Step 19:

How do I do the calculations?

Answer: (12 * 2 * 17) / (28) = 14.6

 12 1 2 NH3 17 g = 14.6g NH3 28 1 1

ARE WE DONE YET? NNNOOOOOO!!!! We have only one answer (14.6 g NH3) of 2 that we need. Now we have to start all over again with the next piece of information 3g of H2 and end our conversion at the same place.

Step 20:

How do we set up the second set of conversions for this problem?

 3 g H2 g NH3 1

Step 21:

What is the first conversion?

Answer: molar mass (grams to mole ratio) of H2 found on the periodic table

Step 22:

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 3 g H2 mol g NH3 g

Step 23:

What is the next step?

Answer: Fill in the numbers and cross out units

 3 g H2 1 mol g NH3 2 g

Step 24:  Simplify

 3 H2 1 mol g NH3 2

Step 25:

What is the next conversion?

Answer: mole to mole ratio (coefficient ratio)

Step 26:

How do I set it up?

Answer: units first, set up the units that you need to cancel out (in red)

 3 H2 1 mol NH3 g NH3 2 H2

Step 27:

What is the next step?

Answer: Fill in the numbers and cross out units

 3 H2 1 mol 2 NH3 g NH3 2 3 H2

Step 28:  Simplify

 3 1 mol 2 NH3 g NH3 2 3

Step 29:

What is the first conversion?

Answer: molar mass (grams to mole ratio) of NH3 found on the periodic table

Step 30:

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 3 H2 1 mol 2 NH3 g = g NH3 2 3 H2 mol

Step 31:

What is the next step?

Answer: Fill in the numbers and cross out units

 3 1 mol 2 NH3 17 g = g NH3 2 3 1 mol

Step 32:  Simplify

 3 1 2 NH3 17 g = g NH3 2 3 1

Step 33:

How do I know I am done with this conversion?

Answer: The only units left are the units that match the answer. In this case g and NH3

 3 1 2 NH3 17 g = g NH3 2 3 1

Step 34:

How do I do the calculations?

Answer: (3 * 2 * 17) / (2 * 3) = 17

 3 1 2 NH3 17 g = 17 g NH3 2 3 1

Step 35:

NOW WE HAVE TWO ANSWERS? SO WHAT DO WE DO WITH THEM? SINCE THEY ARE THE SAME UNITS WE CAN COMPARE THEM. 17 g of NH3 versus 14.6 g NH3. The one that is less is the one we wish to focus on. From the 14.6 g NH3 we tract back to where we calculated that from. It was from the 12 g N2. So this is our limiting reagent. It makes the least amount of product.

Step 36:

COMPLETE ANSWER: Limited reagent is N2 and we can make 14.6 g NH3

VIDEO Stoichiometry (Limited Reactant) Conversions Demonstrated Example 7: If you have 4 g of Ba(OH)2 and 2 g of HCl then what is your excess reactant? How much of the excess reactant will be left over? You will need a periodic table to help solve this problem.

Ba(OH)2(aq) + 2 HCl(aq) ——> 2 H2O(l) + BaCl2(aq)

Step 1:

What information does the question supply us with?

Answer: 4 g of Ba(OH)2 and 2 g of HCl

Step 2:

What units does the question ask?

Answer: the excess reactant and how much is left over after the reaction occurs

Step 3:

Where do we start?

Answer: With excess reactant problems it is harder to figure out where to start. Basically you want to do a comparison between the two reactants so you want to convert one of them into the units of the other. You can start with either of the pieces of information the questions supplies us with. I will start with 4 g of Ba(OH)2 and convert to g HCl.

Step 4:

How many conversions must we do?

Answer: Look at the conversion map. We pass through 3 arrows when we go from Grams of A —> moles of A —> moles of B —> Grams of B. 3 arrows = 3 conversions

Step 5:

How do we set up the first set of conversions for this problem?

 4 g Ba(OH)2 g HCl 1

Step 6:

What is the first conversion?

Answer: molar mass (grams to mole ratio) of Ba(OH)2 found on the periodic table

Step 7:

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 4 g Ba(OH)2 mol g HCl g

Step 8:

What is the next step?

Answer: Fill in the numbers and cross out units

 4 g Ba(OH)2 1 mol g HCl 171 g

Step 9:  Simplify

 4 Ba(OH)2 1 mol g HCl 171

Step 10:

What is the next conversion?

Answer: mole to mole ratio (coefficient ratio)

Step 11:

How do I set it up?

Answer: units first, set up the units that you need to cancel out (in red)

 4 Ba(OH)2 1 mol HCl g HCl 171 Ba(OH)2

Step 12:

What is the next step?

Answer: Fill in the numbers and cross out units

 4 Ba(OH)2 1 mol 2 HCl g HCl 171 1 Ba(OH)2

Step 13:  Simplify

 4 1 mol 2 HCl g HCl 171 1

Step 14:

What is the next conversion?

Answer: molar mass (grams to mole ratio) of Ba(OH)2 found on the periodic table

Step 15:

How do I put that in?

Answer: units first, set up the units that need to cancel out (in red)

 4 1 mol 2 HCl g = g HCl 171 1 mol

Step 16:

What is the next step?

Answer: Fill in the numbers and cross out units

 4 1 mol 2 HCl 36 g = g HCl 171 1 1 mol

Step 17:  Simplify

 4 1 2 HCl 36 g = g HCl 171 1 1

Step 18:

How do I know I am done with this conversion?

Answer: The only units left are the units that match the answer. In this case g and HCl

 4 1 2 HCl 36 g = g HCl 171 1 1

Step 19:

How do I do the calculations?

Answer: (4 * 2 * 36) / (171) = 1.68

 4 1 2 HCl 36 g = 1.68g HCl 171 1 1

Step 20:

We can now compare this conversion answer of 1.68 g HCl to the original piece of information we have of 2 g of HCl. We can only compare these because they have the same units. Since 2 g HCl is higher than the 1.68 g of HCl the HCl is the excess reactant. How much of the excess reactant is left over? For that we can take the difference between the two numbers.

2 g HCl – 1.68 g HCl = 0.32 g HCl

Step 21:

COMPLETE ANSWER: HCl is the excess reactant and 0.32 g of HCl is left after the reaction.

PRACTICE PROBLEMS: Solve these limited reactant stoichiometry problems. Don’t forget to use the periodic table and the conversion map when you need them.

If you have 15g of C4 and 20g of O2 then what is your limited reactant? How many grams of CO2 can you make?

C4(s) + 4 O2(g) —-> 4 CO2(g)

Answer: O2 is the limiting reactant and 27.5 g of CO2 can be made

If you have 7 g of CO32- and 6 g of Fe3+ then what is your limited reactant? How many grams of Fe2(CO3)3 can you make?

2 Fe3+(aq) + 3 CO32-(aq) —-> Fe2(CO3)3(s)

Answer: CO32- is the limiting reactant and 11.36 g of Fe2(CO3)3 can be made

If you have 0.6 g of MgBr2 and 0.9 g of NaI then what is your excess reactant? How much of the excess reactant will be left over?

MgBr2(aq) + 2 NaI(aq) ——> MgI2(aq) + 2 NaBr(aq)

Answer: MgBr2 is your excess reactant and 0.078 g of MgBr2 is left after the reaction

If you have 18 g of C4H10 and 30 g of O2 then what is your excess reactant? How much of the excess reactant will be left over?

2 C4H10 (l) + 13 O2 (g) ——> 8 CO2 (g) + 10 H2O(g)

Answer: C4H10 is the excess reactant and 99g of C4H10 is left after the reaction.  