1. calculate the actual yield of magnesium oxide for each trial.

trial 1:
trial 2:

2. magnesium is the limiting reactant in this experiment. calculate the theoretical yield of mgo for each trial.

trial 1:
trial 2:

3. determine the percent yield of mgo for your experiment for each trial.

trial 1:
trial 2:

4. determine the average percent yield of mgo for the two trials.

Answer 1 : The actual yield of magnesium oxide for trial 1 and trial 2 are, 0.593 g and 0.586 g respectively.

Solution :

Answer 2 : The theoretical yield of MgO for trial 1 and trial 2 are, 0.6 g and 0.596 g respectively.

Solution :

First we have to calculate the mass of Mg for trial 1 and trial 2.

Now we have to calculate the moles of Mg for trial 1 and trial 2

Now we have to calculate the moles of for trial 1 and trial 2.

The given balanced chemical reaction is,

From the balanced reaction, we conclude that

For trial 1 :

As, 2 moles of Mg react to give 2 moles of

So, 0.015 moles of Mg react to give 0.015 moles of

For trial 2 :

As, 2 moles of Mg react to give 2 moles of

So, 0.0149 moles of Mg react to give 0.0149 moles of

Now we have to calculate the mass or theoretical yield of for trial 1 and trial 2.

Answer 3 : The percent yield of MgO for trial 1 and trial 2 are, 98.8% and 98.3% respectively.

Solution :

Answer 4 : The average percent yield of MgO for the two trials is, 98.55%

Solution :

Explanation:

We are given:

Mass of empty crucible with lid = 26.698 g

Mass of Mg O + crucible + lid = 27.291 g

Actual yield of MgO = (27.291 - 26.698) g = 0.593 g

Mass of empty crucible with lid = 26.687 g

Mass of Mg O + crucible + lid = 27.273 g

Actual yield of MgO = (27.273 - 26.687) g = 0.586 g

To calculate the theoretical yield of magnesium oxide, we first need to find the actual yield of magnesium for both the trials.

Mass of empty crucible with lid = 26.698 g

Mass of Mg + crucible + lid = 27.060 g

Actual yield of Mg = (27.060 - 26.698) g = 0.362 g

Mass of empty crucible with lid = 26.698 g

Mass of Mg + crucible + lid = 27.046 g

Actual yield of Mg = (27.046 - 26.698) g = 0.359 g

Now, we calculate the number of moles of magnesium for both the trials.

Molar mass of magnesium = 24.31 g/mol

Then, we calculate the number of moles of MgO for both the trial.

The balanced chemical equation for the formation of magnesium oxide follows:

By Stoichiometry of the reaction:

2 moles of Mg metal produces 2 moles of MgO

So, 0.015 moles of Mg will produce = moles of MgO.

By Stoichiometry of the reaction:

2 moles of Mg metal produces 2 moles of MgO

So, 0.0148 moles of Mg will produce = moles of MgO.

Now, calculating the theoretical yield of magnesium oxide.

To calculate the percentage yield, we use the formula:

Actual yield of MgO = 0.593g

Theoretical yield of MgO = 0.6045 g

Putting values in above equation, we get:

Actual yield of MgO = 0.586 g

Theoretical yield of MgO = 0.5964 g

Putting values in above equation, we get:

To calculate the average yield, we add on total number of yields and divide it by the number of yields.

the answer is definitely c.