Question

post - lab questions: 1. write the balanced chemical equation for the reaction of magnesium metal and oxygen gas. 2. calculate the theoretical yield of the magnesium oxide in your experiment. show all you work to receive credit. 3. calculate the percent yield. show all work. 4. a student forgets to sand the magnesium metal before the experiment. would his percent yield be higher or lower than expected? explain. 5. a student does not add water to the mixture to remove the magnesium nitride (mg₃n₂). would his percent yield be higher or lower than expected? explain.

Explanation:

Step1: Write balanced chemical equation

The reaction of magnesium (Mg) and oxygen ($O_2$) is $2Mg + O_2
ightarrow2MgO$.

Step2: Calculate theoretical yield (assuming initial amount of Mg given)

Let's assume the mass of magnesium used is $m_{Mg}$ grams. The molar mass of Mg is $M_{Mg}=24.31\ g/mol$, and the molar mass of MgO is $M_{MgO}=40.31\ g/mol$. First, find the number of moles of Mg, $n_{Mg}=\frac{m_{Mg}}{M_{Mg}}$. From the balanced - equation, the mole ratio of Mg to MgO is 1:1, so the number of moles of MgO produced theoretically, $n_{MgO}=n_{Mg}$. Then the theoretical mass of MgO, $m_{theo - MgO}=n_{MgO}\times M_{MgO}=\frac{m_{Mg}}{24.31}\times40.31$.

Step3: Calculate percent yield (assuming actual yield given)

Let the actual yield of MgO be $m_{actual - MgO}$ grams. The percent yield is calculated using the formula $\text{Percent Yield}=\frac{m_{actual - MgO}}{m_{theo - MgO}}\times100\%$.

Step4: Analyze effect of not sanding Mg

If a student forgets to sand the magnesium metal before the experiment, there is a layer of magnesium oxide on the surface of Mg. This means that the actual amount of reactive Mg is less than the measured amount. So, the amount of MgO produced will be less than expected, and the percent yield will be lower.

Step5: Analyze effect of not removing $Mg_3N_2$

If a student does not add water to the mixture to remove the magnesium nitride ($Mg_3N_2$), the mass of the final product will be higher because $Mg_3N_2$ is present in the product mixture. Since the percent - yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$, and the actual yield appears higher due to the presence of $Mg_3N_2$, the percent yield will be higher.

Answer:

1. $2Mg + O_2

ightarrow2MgO$

1. Theoretical yield calculation depends on the initial mass of Mg. If mass of Mg is $m_{Mg}$ grams, $m_{theo - MgO}=\frac{m_{Mg}}{24.31}\times40.31$ grams.
2. Percent yield = $\frac{m_{actual - MgO}}{m_{theo - MgO}}\times100\%$ (where $m_{actual - MgO}$ is the actual mass of MgO obtained).
3. Lower. Because there is a non - reactive layer of MgO on the surface of Mg reducing the amount of reactive Mg.
4. Higher. Because the presence of $Mg_3N_2$ in the product mixture increases the measured actual yield.