Solving Limited Reactants Problems: Help Needed

  • Thread starter bluegirlbalance
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In summary, the conversation discussed limited reactants and how to calculate the maximum amount of product that can be made based on the available reactants. An example of making fake blood with limited ingredients was given, and equations were provided for calculating the maximum amount of product that can be formed. The conversation also emphasized the importance of using moles in these calculations.
  • #1
bluegirlbalance
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I totally need help on these problems about limited reactants. My teacher gave me an example: You intend to use the following recipe (Caution: it will turn your mouth bright red—be careful):

1 can frozen white grape juice concentrate

2 Tbs. red food color

1 tsp. yellow food color

Suppose for your movie you need to make 10 batches of this fake blood. In your freezer you have 9 cans of white grape juice, you have 15 tablespoons of red food color and you have 8 teaspoons of yellow food color. Obviously, you need to go to the store, but how many batches can you make?

9 cans of white grape juice will make 9 batches.

15 tbs. of red food color can make 7.5 batches (2 tbs. each)

8 tsp. of yellow food color can make 8 batches.

So...The answer must be 7.5 batches if you want to make half a batch.



Then she gave me these equations to solve:
1. How many grams SO2 can be formed from 20.0 g of S and 160g O2?
2. How many gramsSO2 can be formed from 20 g S and 15.0 g of O2?

(Plese show the balanced equation and your work.)

The balanced equation is S + O2 ---> SO2. What I don't know how to do is the work to figure out the equations above.
 
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  • #2
This is why measurement in moles is so useful. You can come up with certain equivalencies from your balanced equation:

1 mol S can yield 1 mol SO2
1 mol O2 can yield 1 mol SO2

Then convert your given amounts of S and O2 from grams to moles. You can then use those equivalencies. I won't solve your problem but I'll show you how to do a similar problem:

2H2 + O2 --> 2H2O

The atomic weight of H is 1 g/mol, so the molecular weight of H2 is 2 g/mol. The atomic weight of O is 16 g/mol, so the molecular weight of O2, is 32 g/mol.

Now our equivalencies (which we get from the balanced equation above):

2 mol H2 = 2 mol H2O
1 mol O2 = 2 mol H2O

Say we are given 14 g H2 and 50 g O2. Convert these amounts to moles:

[tex]
\frac{14\,g\,H_2}{1} \frac{1\,mol\,H_2}{2\,g\,H_2} = 7\,mol\,H_2
[/tex]

[tex]
\frac{50\,g\,O_2}{1} \frac{1\,mol\,O_2}{32\,g\,O_2} = 1.5625\,mol\,O_2
[/tex]

Now use the equivalencies we have from the reaction equation to see how much of the product will be made given our reactants:

[tex]
\frac{7\,mol\,H_2}{1} \frac{2\,mol\,H_2 O}{2\,mol\,H_2} = 7\,mol\,H_2 O
[/tex]

[tex]
\frac{1.5625\,mol\,O_2}{1} \frac{2\,mol\,H_2 O}{1\,mol\,O_2} = 3.125\,mol\,H_2 O
[/tex]

As you can see, given our amount of H2 and O2, we can only make 3.125 mol H2O. O2 limits the reaction since it it all used up when this much H2O has been made, production of H2O stops, and we have some extra H2 left over.
 
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  • #3


I would first start by identifying the limiting reactant in each equation. In order to do this, we need to convert the given masses of S and O2 into moles. We can do this by using their respective molar masses, which can be found on the periodic table. The molar mass of S is 32.06 g/mol and the molar mass of O2 is 32.00 g/mol.

1. For the first equation, we have 20.0 g of S and 160 g of O2. Converting these masses into moles, we get 0.624 mol of S and 5.00 mol of O2. Since the ratio of S to O2 in the balanced equation is 1:1, we can see that there is an excess of O2 (5.00 mol) compared to S (0.624 mol). Therefore, O2 is the limiting reactant and we can only form 0.624 mol or 19.98 g of SO2.

2. For the second equation, we have 20 g of S and 15.0 g of O2. Converting these masses into moles, we get 0.624 mol of S and 0.469 mol of O2. In this case, we can see that there is an excess of S (0.624 mol) compared to O2 (0.469 mol). Therefore, S is the limiting reactant and we can only form 0.469 mol or 15.01 g of SO2.

In order to solve these types of limited reactant problems, it is important to first identify the limiting reactant and then use the mole ratios from the balanced equation to determine the maximum amount of product that can be formed. I hope this helps with your understanding of limited reactant problems. If you need further assistance, I suggest reaching out to your teacher or a tutor for additional help.
 

1. What are limited reactants and why do we need to solve problems related to them?

Limited reactants are the reactants in a chemical reaction that are completely consumed, limiting the amount of product that can be formed. It is important to solve problems related to limited reactants in order to accurately predict the amount of product that will be produced and to determine the most efficient way to carry out a reaction.

2. How do I identify the limiting reactant in a chemical reaction?

To identify the limiting reactant, you must first write out the balanced chemical equation for the reaction. Then, you can compare the number of moles of each reactant present to the stoichiometric ratio in the balanced equation. The reactant with the smaller number of moles is the limiting reactant.

3. What is the difference between theoretical yield and actual yield?

Theoretical yield is the predicted amount of product that should be formed based on the amount of limiting reactant present and the stoichiometric ratio in the balanced equation. Actual yield is the amount of product that is actually obtained from the reaction. The actual yield is often less than the theoretical yield due to factors such as incomplete reactions or side reactions.

4. How do I calculate the percent yield of a reaction?

The percent yield of a reaction is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. This value represents the efficiency of the reaction and can be affected by various factors such as the purity of the reactants and the conditions in which the reaction is carried out.

5. What are some common strategies for solving limited reactants problems?

Some common strategies for solving limited reactants problems include converting all units to moles, identifying the limiting reactant, calculating the theoretical yield, and using the limiting reactant to determine the amount of excess reactant present. It is also important to consider any factors that may affect the percent yield and adjust calculations accordingly.

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