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Determine the percent yield of MgO for your experiment for each trial.

Determine the percent yield of MgO for your experiment for each trial.-example-1
User LambergaR
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1 Answer

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The equation of the reaction of the lab is:

2 Mg + O₂ -----> 2 MgO

We will use this equation to find the percent yield:

% yield = actual yield/theoretical yield * 100

We can calculate from the given data the actual yield of each trial.

Mass of MgO produced in trial 1 = 27.272 g - 26.690 g

Mass of MgO produced in trial 1 = 0.582 g

Actual yield 1 = 0.582 g

Mass of MgO produced in trial 2 = 27.351 g - 26.681 g

Mass of MgO produced in trial 2 =

Actual yield 2 = 0.67 g

Now we know the actual yield of both trials, but we still have to find the theoretical yield.

2 Mg + O₂ -----> 2 MgO

According to this equation, 2 moles of Mg (with excess O₂) will produce 2 moles of MgO, so their relationship is 1 to 1.

To find the theoretical yield we will use that relationship. First we have to find the number of moles of Mg that reacted in each trial.

mass of Mg in trial 1 = 27.044 g - 26.690 g

mass of Mg in trial 1 = 0.354 g

mass of Mg in trial 2 = 27.089 g - 26.681 g

mass of Mg in trial 2 = 0.408 g

molar mass of Mg = 24.31 g/mol

moles of Mg in trial 1 = 0.354 g/(24.31 g/mol)

moles of Mg in trial 1 = 0.014561 moles

moles of Mg in trial 2 = 0.408 g/(24.31 g/mol)

moles of Mg in trial 2 = 0.016783 moles

We said before that 1 mol of Mg will produce 2 moles of MgO. Let's find the number of moles that should have been produced in each trial.

moles of MgO in trial 1 = 0.014561 moles of Mg * 1 mol of MgO/(1 mol of Mg)

moles of MgO in trial 1 = 0.014561 moles

moles of MgO in trial 2 = 0.016783 moles of Mg * 1 mol of MgO/(1 mol of Mg)

moles of MgO in trial 2 = 0.016783 moles

Now we can find the theoretical yield of each trial using the molar mass of MgO:

molar mass of MgO = 24.31 g/mol + 16.00 g/mol

molar mass of MgO = 40.31 g/mol

mass of MgO theoretically produced in trial 1 = 0.014561 moles * 40.31 g/mol

mass of MgO theoretically produced in trial 1 = 0.5870 g

theoretical yield trial 1 = 0.5870 g

mass of MgO theoretically produced in trial 2 = 0.016783 moles * 40.31 g/mol

mass of MgO theoretically produced in trial 2 = 0.6765 g

theoretical yield trial 2 = 0.6765 g

Finally we can find the percent yield for each trial:

% yield trial 1 = actual yield trial 1/theoretical yield trial 1 * 100

% yield trial 1 = 0.582 g/0.5870 g * 100

% yield trial 1 = 99.15 %

% yield trial 2 = actual yield trial 2/theoretical yield trial 2 * 100

% yield trial 2 = 0.67 g/0.6765 g * 100

% yield trial 2 = 99.04 %

Answer: the % yield for trial 1 is 99.15 % and for trial 2 is 99.04 %

User Attaullah
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