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.
5 eggs | 24 cookies | = 60 cookies |
2 eggs |
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 N_{2} and 3g of H_{2} then what is your limited reactant? How many grams of NH_{3} can you make? You will need a periodic table to help solve this problem.
N_{2(g)} + 3 H_{2(g)} —-> 2 NH_{3(g)}
Step 1:
What information does the question supply us with?
Answer: 12 g of N_{2} and 3g of H_{2}
Step 2:
What units does the question ask?
Answer: the limited reactant and g NH_{3 }
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 N_{2}.
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?
Answer:
12 g N_{2} | g NH_{3} | |||
1 |
Step 6:
What is the first conversion?
Answer: molar mass (grams to mole ratio) of N_{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)
12 g N_{2} | mol | g NH_{3} | ||
g |
Step 8:
What is the next step?
Answer: Fill in the numbers and cross out units
12 g N_{2} | 1 mol | g NH_{3} | ||
28 g |
Step 9: Simplify
12 N_{2} | 1 mol | g NH_{3} | ||
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 N_{2} | 1 mol | NH_{3} | g NH_{3} | |
28 | N_{2} |
Step 12:
What is the next step?
Answer: Fill in the numbers and cross out units
12 N_{2} | 1 mol | 2 NH_{3} | g NH_{3} | |
28 | 1 N_{2} |
Step 13: Simplify
12 | 1 mol | 2 NH_{3} | g NH_{3} | |
28 | 1 |
Step 14:
What is the next conversion?
Answer: molar mass (grams to mole ratio) of NH_{3} 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 NH_{3} | g = | g NH_{3} |
28 | 1 | mol |
Step 16:
What is the next step?
Answer: Fill in the numbers and cross out units
12 | 1 mol | 2 NH_{3} | 17 g = | g NH_{3} |
28 | 1 | 1 mol |
Step 17: Simplify
12 | 1 | 2 NH_{3} | 17 g = | g NH_{3} |
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 NH_{3}
12 | 1 | 2 NH_{3} | 17 g = | g NH_{3} |
28 | 1 | 1 |
Step 19:
How do I do the calculations?
Answer: (12 * 2 * 17) / (28) = 14.6
12 | 1 | 2 NH_{3} | 17 g = | 14.6g NH_{3} |
28 | 1 | 1 |
ARE WE DONE YET? NNNOOOOOO!!!! We have only one answer (14.6 g NH_{3}) of 2 that we need. Now we have to start all over again with the next piece of information 3g of H_{2 }and end our conversion at the same place.
Step 20:
How do we set up the second set of conversions for this problem?
Answer:
3 g H_{2} | g NH_{3} | |||
1 |
Step 21:
What is the first conversion?
Answer: molar mass (grams to mole ratio) of H_{2} 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 H_{2} | mol | g NH_{3} | ||
g |
Step 23:
What is the next step?
Answer: Fill in the numbers and cross out units
3 g H_{2} | 1 mol | g NH_{3} | ||
2 g |
Step 24: Simplify
3 H_{2} | 1 mol | g NH_{3} | ||
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 H_{2} | 1 mol | NH_{3} | g NH_{3} | |
2 | H_{2} |
Step 27:
What is the next step?
Answer: Fill in the numbers and cross out units
3 H_{2} | 1 mol | 2 NH_{3} | g NH_{3} | |
2 | 3 H_{2} |
Step 28: Simplify
3 | 1 mol | 2 NH_{3} | g NH_{3} | |
2 | 3 |
Step 29:
What is the first conversion?
Answer: molar mass (grams to mole ratio) of NH_{3} 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 H_{2} | 1 mol | 2 NH_{3} | g = | g NH_{3} |
2 | 3 H_{2} | mol |
Step 31:
What is the next step?
Answer: Fill in the numbers and cross out units
3 | 1 mol | 2 NH_{3} | 17 g = | g NH_{3} |
2 | 3 | 1 mol |
Step 32: Simplify
3 | 1 | 2 NH_{3} | 17 g = | g NH_{3} |
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 NH_{3}
3 | 1 | 2 NH_{3} | 17 g = | g NH_{3} |
2 | 3 | 1 |
Step 34:
How do I do the calculations?
Answer: (3 * 2 * 17) / (2 * 3) = 17
3 | 1 | 2 NH_{3} | 17 g = | 17 g NH_{3} |
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 NH_{3} versus 14.6 g NH_{3}. 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 N_{2}. So this is our limiting reagent. It makes the least amount of product.
Step 36:
COMPLETE ANSWER: Limited reagent is N_{2} and we can make 14.6 g NH_{3}
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 H_{2}O_{(l)} + BaCl_{2(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?
Answer:
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 C_{4} and 20g of O_{2} then what is your limited reactant? How many grams of CO_{2} can you make?
C_{4(s)} + 4 O_{2(g)} —-> 4 CO_{2(g)}
Answer: O_{2} is the limiting reactant and 27.5 g of CO_{2} can be made
If you have 7 g of CO_{3}^{2-} and 6 g of Fe^{3+} then what is your limited reactant? How many grams of Fe_{2}(CO_{3})_{3 }can you make?
2 Fe^{3+}_{(aq)} + 3 CO_{3}^{2-}_{(aq)} —-> Fe_{2}(CO_{3})_{3(s)}
Answer: CO_{3}^{2-} is the limiting reactant and 11.36 g of Fe_{2}(CO_{3})_{3} can be made
If you have 0.6 g of MgBr_{2} and 0.9 g of NaI then what is your excess reactant? How much of the excess reactant will be left over?
MgBr_{2(aq)} + 2 NaI_{(aq)} ——> MgI_{2(aq)} + 2 NaBr_{(aq)}
Answer: MgBr_{2} is your excess reactant and 0.078 g of MgBr_{2} is left after the reaction
If you have 18 g of C_{4}H_{10} and 30 g of O_{2} then what is your excess reactant? How much of the excess reactant will be left over?
2 C_{4}H_{10 (l)} + 13 O_{2 (g)} ——> 8 CO_{2 (g)} + 10 H_{2}O_{(g) }
Answer: C_{4}H_{10} is the excess reactant and 99g of C_{4}H_{10 }is left after the reaction.