Magnesium oxide can be made by heating magnesium metal in
the pre...

Question

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. Determine the theoretical yield. Determine the percent yield for the reaction.

Accepted Answer

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

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