Question

Percent Yield Lab Report

Answer 1

Based on the data obtained from the reaction, the following conclusions can be made;

• according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
• the percent yield is less than 100% because of loss in mass of either reactants or products.
• the errors could have occurred during the weighing and transfer of reactants and products
• repeated measurements are required in order to improve accuracy

What is the percent yield of the reaction?

Equation of the reaction is given below:

• 2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid  = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

Moles of Mg used

moles of Mg = mass/molar mass

• molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

Moles of MgO expected

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced =  mass/molar mass

• molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

Percent yield

• Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

Moles of Mg used

moles of Mg = mass/molar mass

• molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

Moles of MgO expected

Based on the equation of reaction;

moles of MgO expected = 0.0170

• moles of MgO produced =  mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

Percent yield

• Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

• according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
• the percent yield is less than 100% because of loss in mass of either reactants or products.
• the errors could have occurred during the weighing and transfer of reactants and products
• repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546