Chemistry, 10.11.2019 17:31 shortty1111

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.

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

Answers: 3

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Answer from: 00manray

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

Solution :

$\text{Actual yield of MgO for trial 1}=\text{Mass of MgO, crucible & lid}-\text{Mass of empty crucible with lid}=27.291g-26.698g=0.593g$

$\text{Actual yield of MgO for trial 2}=\text{Mass of MgO, crucible & lid}-\text{Mass of empty crucible with lid}=27.273g-26.687g=0.586g$

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.

$\text{Mass of Mg for trial 1}=\text{Mass of Mg,crucible & lid}-\text{Mass of empty crucible with lid}=27.060g-26.698g=0.362g$

$\text{Mass of Mg for trial 2}=\text{Mass of Mg,crucible & lid}-\text{Mass of empty crucible with lid}=27.046g-26.687g=0.359g$

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

$\text{Moles of Mg for trial 1}=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{0.362g}{40g/mole}=0.015moles$

$\text{Moles of Mg for trial 2}=\frac{\text{Mass of }Mg}{\text{Molar mass of }Mg}=\frac{0.359g}{40g/mole}=0.0149moles$

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

The given balanced chemical reaction is,

$2Mg+O_2\rightarrow 2MgO$

From the balanced reaction, we conclude that

For trial 1 :

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

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

For trial 2 :

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

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

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

$\text{Mass of MgO for trial 1}=\text{Moles of }MgO\times \text{Molar mass of }MgO=(0.015mole)\times (40g/mole)=0.6g$

$\text{Mass of MgO for trial 2}=\text{Moles of }MgO\times \text{Molar mass of }MgO=(0.0149mole)\times (40g/mole)=0.596g$

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

Solution :

$\%\text{ yield of MgO for trial 1}=\frac{\text{Actual yield of MgO}}{\text{Theoretical yield of MgO}}\times 100=\frac{0.593g}{0.6g}\times 100=98.8\%$

$\%\text{ yield of MgO for trial 2}=\frac{\text{Actual yield of MgO}}{\text{Theoretical yield of MgO}}\times 100=\frac{0.586g}{0.596g}\times 100=98.3\%$

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

Solution :

$\text{Average percent yield of MgO}=\frac{\%\text{ yield of MgO for trial 1}+\%\text{ yield of MgO for trial 2}}{2}=\frac{98.8+98.3}{2}=98.55\%$

Answer from: Lilyy1k

Explanation:

For 1:

We are given:

Trial 1:

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

Trial 2:

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

For 2:

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

Trial 1:

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

Trial 2:

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

Trial 1:

$\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=\frac{0.362g}{24.31g/mole}=0.015moles$

Trial 2:

$\text{Moles of Mg}=\frac{\text{Mass of Mg}}{\text{Molar mass of Mg}}=\frac{0.359g}{24.31g/mole}=0.0148moles$

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

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

$2Mg+O_2\rightarrow 2MgO$

Trial 1:

By Stoichiometry of the reaction:

2 moles of Mg metal produces 2 moles of MgO

So, 0.015 moles of Mg will produce = $\frac{2}{2}\times 0.015=0.015$ moles of MgO.

Trial 2:

By Stoichiometry of the reaction:

2 moles of Mg metal produces 2 moles of MgO

So, 0.0148 moles of Mg will produce = $\frac{2}{2}\times 0.0148=0.0148$ moles of MgO.

Now, calculating the theoretical yield of magnesium oxide.

Trail 1:

$\text{Theoretical yield of MgO}=\text{Moles of MgO}\times \text{Molar mass of MgO}$

$\text{Theoretical yield of MgO}=(0.015mole)\times (40.3g/mole)=0.6045g$

Trial 2:

$\text{Theoretical yield of MgO}=\text{Moles of MgO}\times \text{Molar mass of MgO}$

$\text{Theoretical yield of MgO}=(0.0148mole)\times (40g/mole)=0.5964g$

For 3:

To calculate the percentage yield, we use the formula:

$\% \text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Trial 1:

Actual yield of MgO = 0.593g

Theoretical yield of MgO = 0.6045 g

Putting values in above equation, we get:

$\%\text{ yield of MgO}=\frac{0.593g}{0.6045g}\times 100=98.09\%$

Trial 2:

Actual yield of MgO = 0.586 g

Theoretical yield of MgO = 0.5964 g

Putting values in above equation, we get:

$\%\text{ yield of MgO}=\frac{0.586g}{0.5964g}\times 100=98.25\%$

For 4:

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

$\text{Average percent yield of MgO}=\frac{\%\text{ yield of MgO for trial 1}+\%\text{ yield of MgO for trial 2}}{2}$

$\text{Average percent yield of MgO}=\frac{98.09+98.3}{2}=98.195\%$

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

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