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.

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

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

Answer from: Quest

mine i add some sprinkles to it

oh and its pronounced asscream

not ice

Answer from: Quest

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

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