Answer:
Part 1
Theoretical yield of MgO for trial 1 = 0.84 g
Theoretical yield of MgO for trial 2 = 1.01 g
Part 2
Percent yield trial 1 = 28.6 %
Percent yield trial 2 = 49.9 %
Part 3
Average percent yield of MgO for two trial = 39.25 %
Step-by-step explanation:
Part 1.
Data Given
Trial 1 Trial 2
mass of empty crucible and lid: 26.679 g 26.685 g
mass of Mg metal, crucible and lid: 26.931 g 26.988 g
mass of MgO, crucible and lid: 27.090 g 27.179 g
Theoretical yield of MgO for trial 1 and 2 = ?
Solution:
As Mg is limiting reagent so amount of MgO depends on the amount of Mg.
So, now we will look for the reaction to calculate theoretical yield
MgO form by the following reaction:
Mg + O₂ ---------> 2 MgO
1 mol 2 mol
Convert moles to mass
Molar mass of Mg = 24 g/mol
Molar mass of MgO = 24 + 16 = 40 g/mol
So,
Mg + O₂ ---------> 2 MgO
1 mol (24 g/mol) 2 mol(40 g/mol)
24 g 80 g
So,
24 g of Mg gives 80 g of MgO
To Calculate theoretical yield of MgO for Trial 1
First we look for the mass of Mg in the Crucible
Mass of Mg = mass of Mg metal, crucible and lid - mass of empty crucible and lid
Mass of Mg = 26.931 g - 26.679 g
Mass of Mg = 0.252 g
As we come to know that 24 g of Mg gives 80 g of MgO, then amount of Mg from trial 1 that is 0.252 g will produce how many grams of MgO
Apply unity formula
24 g of Mg ≅ 80 g of MgO
0.252 g of Mg ≅ X g of MgO
Do cross multiplication
X g of MgO = 0.252 g x 80 g / 24 g
X g of MgO = 0.84 g
So the theoretical yield of MgO is 0.84 g
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To Calculate theoretical yield of MgO for Trial 2
First we look for the mass of Mg in the Crucible
Mass of Mg = mass of Mg metal, crucible and lid - mass of empty crucible and lid
Mass of Mg = 26.988 g - 26.685 g
Mass of Mg = 0.303 g
As we come to know that 24 g of Mg gives 80 g of MgO, then amount of Mg from trial 2 that is 0.303 g will produce how many grams of MgO
Apply unity formula
24 g of Mg ≅ 80 g of MgO
0.303 g of Mg ≅ X g of MgO
Do cross multiplication
X g of MgO = 0.303 g x 80 g / 24 g
X g of MgO = 1.01 g
So the theoretical yield of MgO is 1.01 g
__________________________
Part 2
percent yield of MgO for trial 1 and 2 = ?
Solution:
For trial 1
To calculate percent yield we have to know about actual yield of MgO
Mass of MgO = mass of MgO, crucible and lid - mass of empty crucible and lid
Mass of MgO = 27.090 g - 26.685 g
Mass of MgO = 0.24 g
And we also know that
Theoretical yield of MgO for trial 1 = 0.84 g
Formula used
Percent yield = actual yield / theoretical yield x 100
put values in above formula
Percent yield = 0.24 g / 0.84 g x 100
Percent yield = 28.6 %
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For trial 2
To calculate percent yield we have to know about actual yield of MgO
Mass of MgO = mass of MgO, crucible and lid - mass of empty crucible and lid
Mass of MgO = 27.179 g - 26.685 g
Mass of MgO = 0.494 g
And we also know that
Theoretical yield of MgO for trial 2 = 1.01 g
Formula used
Percent yield = actual yield / theoretical yield x 100
put values in above formula
Percent yield = 0.494 g/ 1.01 g x 100
Percent yield = 49.9 %
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Part 3
average percent yield of MgO for the two trials =?
Solution:
As we know
Percent yield trial 2 = 28.6 %
Percent yield trial 2 = 49.9 %
Formula used
Average percent yield = percent yield trial 1 + percent yield trial 2 / 2
Put values in above formula
Average percent yield = 28.6 + 49.9 / 2
Average percent yield = 78.5 / 2
Average percent yield = 39.25 %
Average percent yield of MgO for two trial = 39.25 %