Math Problem Statement

Determine the theoretical yield (maximum amount of product that can be formed) in grams when 0.486 grams of magnesium are reacted with excess pure oxygen. Hint: Use the balanced equation provided in the experiment.

Solution

To determine the theoretical yield, we'll use the reaction between magnesium (Mg) and oxygen (O₂) to form magnesium oxide (MgO). The balanced chemical equation is:

2Mg+O22MgO2 \, \text{Mg} + O_2 \rightarrow 2 \, \text{MgO}

Step 1: Calculate the moles of magnesium.

The molar mass of magnesium (Mg) is 24.305 g/mol. We can calculate the moles of magnesium used:

Moles of Mg=Mass of MgMolar mass of Mg=0.486g24.305g/mol0.020mol of Mg\text{Moles of Mg} = \frac{\text{Mass of Mg}}{\text{Molar mass of Mg}} = \frac{0.486 \, \text{g}}{24.305 \, \text{g/mol}} \approx 0.020 \, \text{mol of Mg}

Step 2: Use the stoichiometry of the reaction.

From the balanced equation, we know that 2 moles of magnesium react with 1 mole of oxygen to produce 2 moles of magnesium oxide. Therefore, the number of moles of magnesium oxide produced is equal to the number of moles of magnesium reacted:

Moles of MgO=0.020mol of Mg\text{Moles of MgO} = 0.020 \, \text{mol of Mg}

Step 3: Calculate the mass of magnesium oxide (MgO).

The molar mass of magnesium oxide (MgO) is:

Molar mass of MgO=Molar mass of Mg+Molar mass of O=24.305g/mol+16.00g/mol=40.305g/mol\text{Molar mass of MgO} = \text{Molar mass of Mg} + \text{Molar mass of O} = 24.305 \, \text{g/mol} + 16.00 \, \text{g/mol} = 40.305 \, \text{g/mol}

Now, we calculate the mass of MgO produced:

Mass of MgO=Moles of MgO×Molar mass of MgO=0.020mol×40.305g/mol=0.806g\text{Mass of MgO} = \text{Moles of MgO} \times \text{Molar mass of MgO} = 0.020 \, \text{mol} \times 40.305 \, \text{g/mol} = 0.806 \, \text{g}

Theoretical Yield:

The theoretical yield of magnesium oxide is 0.806 grams.

Would you like more details or have any questions?


Here are 5 related questions to deepen your understanding:

  1. What is the limiting reactant in a chemical reaction, and how does it affect the theoretical yield?
  2. How would you determine the actual yield from an experiment?
  3. Why is it important to have a balanced chemical equation when calculating theoretical yield?
  4. How would the theoretical yield change if 1.0 grams of magnesium were used instead of 0.486 grams?
  5. What are some factors that can cause the actual yield to be lower than the theoretical yield?

Tip: Theoretical yield assumes perfect conditions. Always compare it with actual yield to calculate percent yield!

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Math Problem Analysis

Mathematical Concepts

Stoichiometry
Mole Concept
Chemical Reactions

Formulas

Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of MgO = Moles of Mg (from balanced equation)
Mass of MgO = Moles of MgO * Molar mass of MgO

Theorems

Law of Conservation of Mass

Suitable Grade Level

High School Chemistry (Grades 9-12)