Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2 Mg(s) + O 2 (g) → 2 MgO(s) When 10.1 g of Mg reacts with 10.5 g O 2 , 11.9 g MgO is collected. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Hey everyone. Our question here states according to the following reaction, we have 283 mg of calcium reacting with mg of oxygen gas. We produce 185 mg of calcium oxide. And they want us to determine the limiting reactant theoretical yield and our percent yield first. In order to determine our limiting reactant, we want to go from the grams of reactant, two moles of reactant and finally into mole of our product, which is calcium oxide. Starting off with our calcium, We have 283 mg of calcium and we want to convert this into grams and we know that we have 10 to the third milligrams per one g. And looking at our periodic table, we find that the molar mass of calcium is 40.07, 8 g per one mole. In order to relate this to calcium oxide. We'll go ahead and take a look at our equation and we see that we have two moles of calcium per two moles of calcium oxide. And when we calculate this out, We end up with a value of 7. Times 10 to the -3 mole of calcium oxide. Now let's go ahead and take a look at oxygen gas. We started off with 367 mg of oxygen gas. Again, we want to convert this into grams And we know that we have 10 to 3rd mg per one g. And looking at our periodic table, We see that we have 32.0 g of oxygen gas per one mole. And from our reaction, we see that we have one mole of oxygen gas per two mol of calcium oxide. And when we calculate this out, we end up with a value of 2.29, 3 times 10 To the -2 mol calcium oxide. Now, when we compare these two, we can see that calcium produce the least amount of calcium oxide. And so this is actually going to be our limiting reacting. Now let's go ahead and determine our theoretical yield to determine our theoretical yield. We're going to need to use the value that we just calculated which is going to be Our 7.06 Times 10 to the -3 mole of calcium oxide. And the reason we chose this value is because calcium are limiting reactant produce this amount. Now we want to convert this into grams of calcium oxide. So we know that per one mole of calcium oxide. When we calculate our molar mass, We find that the molar mass of calcium oxide is 56.0774 g per one mole. And we can go ahead and convert this into milligrams. Since we started off with milligrams. So again, we know that we have one g and this contains 10 to the 3rd Millg. So when we calculate this out, we end up with a value of milligrams of calcium oxide. And this is going to be our theoretical yield. Finally calculating our percent yield. We're going to take our 185 mg, which is our value of calcium oxide that was presented to us in our questions them. So taking our 185 mg of calcium oxide, We're going to divide this by our theoretical yield of 395 mg of calcium oxide. And we want to multiply this by 100%. Since we're calculating for a percentage, This will get us a value of 46.7%. So our final answers here are going to be 46.7% as our percent yield. 395 mg of calcium oxide as our theoretical yield and our limiting reactant is going to be our calcium. So I hope this made sense. And let us know if you have any questions

Ch.4 - Chemical Reactions and Chemical Quantities

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Tro 5th Edition

Ch.4 - Chemical Reactions and Chemical Quantities

Problem 48

Chapter 4, Problem 48

Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2 Mg(s) + O₂(g) → 2 MgO(s) When 10.1 g of Mg reacts with 10.5 g O₂, 11.9 g MgO is collected. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

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1. First, calculate the number of moles of Mg and O2 using their respective molar masses. The molar mass of Mg is approximately 24.31 g/mol and the molar mass of O2 is approximately 32.00 g/mol. Use the formula: number of moles = mass / molar mass.

2. Next, determine the limiting reactant. This is the reactant that will be completely consumed first in the reaction. According to the balanced chemical equation, the mole ratio of Mg to O2 is 2:1. Compare the mole ratio of the reactants to the stoichiometric ratio to determine the limiting reactant.

3. Once the limiting reactant is determined, calculate the theoretical yield of MgO. The theoretical yield is the maximum amount of product that can be formed from the limiting reactant. Use the stoichiometric ratio from the balanced chemical equation to calculate the theoretical yield.

4. The actual yield is the amount of product that was actually formed in the reaction. In this case, the actual yield is given as 11.9 g of MgO.

5. Finally, calculate the percent yield using the formula: percent yield = (actual yield / theoretical yield) * 100%. This gives the efficiency of the reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. To identify the limiting reactant, one must compare the mole ratios of the reactants based on the balanced chemical equation. The reactant that produces the lesser amount of product is the limiting reactant.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactants, as calculated from the balanced chemical equation. It assumes complete conversion of the limiting reactant into product without any losses. This value is essential for calculating the efficiency of a reaction and is typically expressed in grams or moles.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated by comparing the actual yield of a product obtained from the experiment to the theoretical yield. It is expressed as a percentage using the formula: (actual yield / theoretical yield) × 100. A high percent yield indicates a successful reaction, while a low percent yield may suggest incomplete reactions or losses during the process.

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