Question

must calculate the theoretical yield and the average percent yield and average percent yeiled of MgO in the following problem.2Mg + O2 --> 2MgOmass of magnesium for the trial: .326mass of magnesium oxide for the trial 27.198Mg is the limiting reactant.

Answer 1

So,

Given the reaction:

The initial mass of Mg was 0.326, and, in the real experiment, we obtained 0.528 grams of MgO. Right?

Now, let's find the amount of MgO supposed to obtain according to the chemical equation:

Now, the amount that we were supposed to obtain, was 0.54g of MgO. (The result of mutiplying all the previous operations). This is the theoretical yield, what we obtained using the theory.

Now, as you discovered, the mass of MgO in the laboratory was 0.528g so that's the actual yield. This is, what you've found using the experimental process.

Finally, the percent yield can be found using the following equation:

And, we know both values, so let's just replace: