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How do you calculate a limiting reactant?
What is a limiting reactant?
How do you identify limiting and excess reactants?
How do you calculate the theoretical yield of a limiting reactant?
How do you calculate molar mass of a limiting reactant?
What is the difference between limiting reactant and excess reactant?
1) Consider the following reaction: 3 NH4NO3 + Na3PO4 (NH4)3PO4 + 3 NaNO3. Answer the questions above, assuming we started with 30 grams of ammonium nitrate and 50 grams of sodium phosphate.
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May 4, 2024 · Identify the limiting reactant (limiting reagent) in a given chemical reaction. Calculate how much product will be produced from the limiting reactant. Calculate how much reactant(s) remains when the reaction is complete.
What is the limiting reactant if 2.2 g of Mg is reacted with 4.5 L of oxygen at STP? Both of the following give you the same answer. In the first case, you need to do one or the other.
Jul 12, 2023 · The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation.
- Overview
- Limiting reactant and theoretical yield
- Example 1: Using the limiting reactant to calculate theoretical yield
- Step 1: Convert reactant masses to moles
- Step 2: Find the limiting reactant
- Step 3: Calculate the theoretical yield
- Percent yield
- Example 2: Calculating percent yield
- Step 1: Find moles of the limiting reactant
- Step 2: Determine the theoretical yield (in grams)
Limiting reactant and theoretical yield
It’s a classic conundrum: We have five hot dogs and four hot dog buns. How many complete hot dogs can we make?
Assuming that hot dogs and buns combine in a 1:1 ratio, we can make four complete hot dogs. Once we run out of buns, we'll have to stop making complete hot dogs. In other words, the hot dog buns limit the number of complete hot dogs we can produce.
In much the same way, a reactant in a chemical reaction can limit the amounts of products formed by the reaction. When this happens, we refer to the reactant as the limiting reactant (or limiting reagent). The amount of a product that is formed when the limiting reactant is fully consumed in a reaction is known as the theoretical yield. In the case of our hot dog example, we already determined the theoretical yield (four complete hot dogs) based on the number of hot dogs buns we were working with.
Enough about hot dogs, though! In the next example, we'll see how to identify the limiting reactant and calculate the theoretical yield for an actual chemical reaction.
Example 1: Using the limiting reactant to calculate theoretical yield
It’s a classic conundrum: We have five hot dogs and four hot dog buns. How many complete hot dogs can we make?
Assuming that hot dogs and buns combine in a 1:1 ratio, we can make four complete hot dogs. Once we run out of buns, we'll have to stop making complete hot dogs. In other words, the hot dog buns limit the number of complete hot dogs we can produce.
In much the same way, a reactant in a chemical reaction can limit the amounts of products formed by the reaction. When this happens, we refer to the reactant as the limiting reactant (or limiting reagent). The amount of a product that is formed when the limiting reactant is fully consumed in a reaction is known as the theoretical yield. In the case of our hot dog example, we already determined the theoretical yield (four complete hot dogs) based on the number of hot dogs buns we were working with.
Enough about hot dogs, though! In the next example, we'll see how to identify the limiting reactant and calculate the theoretical yield for an actual chemical reaction.
A 2.80 g sample of Al(s) reacts with a 4.15 g sample of ClA2(g) according to the equation shown below.
2Al(s)+3ClA2(g)→2AlClA3(s)
What is the theoretical yield of AlClA3 in this reaction?
To solve this problem, we first need to determine which reactant, Al or ClA2 , is limiting. We can do so by converting both reactant masses to moles and then using one or more mole ratios from the balanced equation to identify the limiting reactant. From there, we can use the amount of the limiting reactant to calculate the theoretical yield of AlClA3 .
Let's start by converting the masses of Al and ClA2 to moles using their molar masses:
2.80g Al×1mol Al26.98g Al=1.04×10−1mol Al4.15g Cl2×1mol Cl270.90g Cl2=5.85×10−2mol Cl2
Now that we know the quantities of Al and ClA2 in moles, we can determine which reactant is limiting. As you'll see below, there are multiple ways to do so, each of which uses the concept of the mole ratio. All of the methods give the same answer, though, so you can choose whichever approach you prefer!
Method 1: For the first method, we'll determine the limiting reactant by comparing the mole ratio between Al and ClA2 in the balanced equation to the mole ratio actually present. In this case, the mole ratio of Al and ClA2 required by balanced equation is
moles of Almoles of Cl2(required)=23=0.66―
and the actual mole ratio is
moles of Almoles of Cl2(actual)=1.04×10−15.85×10−2=1.78
Since the actual ratio is greater than the required ratio, we have more Al than is needed to completely react the ClA2 . This means that the Cl2 must be the limiting reactant. If the actual ratio had been smaller than the required ratio, then we would have had excess Cl2 , instead, and the Al would be limiting.
Our final step is to determine the theoretical yield of AlCl3 in the reaction. Remember that the theoretical yield is the amount of product that is produced when the limiting reactant is fully consumed. In this case, the limiting reactant is ClA2 , so the maximum amount of AlCl3 that can be formed is
5.85×10−2mol Cl2×2mol AlCl33mol Cl2=3.90×10−2mol AlCl3
Note that we had already calculated this value while working through Method 3! Since a theoretical yield is typically reported with units of mass, let's use the molar mass of AlClA3 to convert from moles of AlClA3 to grams:
3.90×10−2mol AlCl3×133.33g AlCl31mol AlCl3=5.20g AlCl3
As we just learned, the theoretical yield is the maximum amount of product that can be formed in a chemical reaction based on the amount of limiting reactant. In practice, however, the actual yield of product—the amount of product that is actually obtained—is almost always lower than the theoretical yield. This can be due to a number of factors, including side reactions (secondary reactions that form undesired products) or purification steps that lower the amount of product isolated after the reaction.
The actual yield of a reaction is typically reported as a percent yield, or the percentage of the theoretical yield that was actually obtained. The percent yield is calculated as follows:
Percent yield=actual yieldtheoretical yield×100%
Based on this definition, we would expect a percent yield to have a value between 0% and 100%. If our percent yield is greater than 100%, that means we probably calculated something incorrectly or made an experimental error. With all this in mind, let's try calculating the percent yield for a precipitation reaction in the following example.
A students mixes 25.0 mL of 0.314M BaClA2 with excess AgNOA3 , causing AgCl to precipitate. The balanced equation for the reaction is shown below.
BaClA2(aq)+2AgNOA3(aq)→2AgCl(s)+Ba(NOA3)A2(aq)
If the student isolates 1.82 g of AgCl(s) , what is the percent yield of the reaction?
To solve this problem, we'll need to use the given information about the limiting reactant, BaClA2 , to calculate the theoretical yield of AgCl for the reaction. Then, we can compare this value to the actual yield of AgCl to determine the percent yield. Let's walk through the steps now:
To determine the theoretical yield of AgCl , we first need to know how many moles of BaClA2 were consumed in the reaction. We're given the volume (0.0250 L ) and molarity (0.314M ) of the BaClA2 solution, so we can find the number of moles of BaClA2 by multiplying these two values:
Moles BaCl2=MBaCl2×liters soln=0.314mol BaCl21L soln×.0250L soln=7.85×10−3mol BaCl2
Now that we have the quantity of BaClA2 in moles, we can find the theoretical yield of AgCl in grams:
7.85×10−3mol BaCl2×2mol AgCl1mol BaCl2×143.32g AgCl1mol AgCl=2.25g AgCl
Aug 14, 2020 · Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation. Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant.
1. What is a limiting reactant? 2. What is an excess reactant? 3. How do you determine the limiting and excess reactant for a reaction? 4. Answer the following questions using the balanced reaction below. CuCl2 + 2 NaNO3 Cu(NO3)2 + 2 NaCl. a.
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