In a reaction, 41.8 g of chromium(III) oxide reacts with 14.8 g of aluminum to produce chromium and aluminum oxide. If 28.6 g of chromium is produced, what mass of aluminum oxide is produced? How do I go about solving this problem?

It's actually easier than it seems. Just requires simple stoichiometry.

Balanced Equation:
Cr₂O₃ + 2 Al -> Al₂O₃ + 2 Cr

28.6 g Cr * (1 mol Cr / 52.00 g Cr) * (1 mol Al₂O₃ / 2 mol Cr) * (101.96 g Al₂O₃/ 1 mol Al₂O₃) = 28.0 g Al₂O₃ (with sig figs)

Well, it seems like you're a fan of chemistry. Let's see if I can cook up a funny solution for you!

To find the mass of aluminum oxide produced, we need to first determine which reactant limits the reaction. We can do this by comparing the moles of each reactant and seeing which one has the lower number of moles.

The molar mass of chromium(III) oxide is approximately 152 g/mol, while the molar mass of aluminum is around 27 g/mol. So, we can start by calculating the number of moles for both reactants.

For chromium(III) oxide, we divide 41.8 g by its molar mass:
41.8 g / 152 g/mol ≈ 0.274 moles

And for aluminum, we divide 14.8 g by its molar mass:
14.8 g / 27 g/mol ≈ 0.548 moles

It seems like chromium(III) oxide is our limiting reactant since it has fewer moles. Now, we can use the balanced equation to find the moles of aluminum oxide produced. According to the equation, one mole of chromium(III) oxide produces one mole of chromium and one mole of aluminum oxide.

Since we produced 0.274 moles of chromium, we should also have 0.274 moles of aluminum oxide. Now we need to convert moles to mass. The molar mass of aluminum oxide is around 102 g/mol, so multiplying the number of moles by the molar mass will give us the mass:

0.274 moles * 102 g/mol ≈ 27.85 g

Therefore, approximately 27.85 g of aluminum oxide should be produced.

And there you have it! The mass of aluminum oxide produced is around 27.85 g. Remember, chemistry can be a reaction without too many complications, just like my attempts at being funny.

To solve this problem, we need to use the concept of stoichiometry, which involves converting the given masses of substances to moles, and then using the balanced equation to determine the moles of the desired substance.

Step 1: Write the balanced chemical equation.
The balanced equation for the reaction between chromium(III) oxide and aluminum is:
2Al + Cr2O3 → 2Cr + Al2O3

Step 2: Calculate the moles of chromium(III) oxide reacted.
Using the molar mass of chromium(III) oxide (Cr2O3), which is 152 g/mol:
Number of moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3
= 41.8 g / 152 g/mol
= 0.274 moles

Step 3: Calculate the moles of aluminum reacted.
Using the molar mass of aluminum (Al), which is 27 g/mol:
Number of moles of Al = Mass of Al / Molar mass of Al
= 14.8 g / 27 g/mol
= 0.548 moles

Step 4: Determine the limiting reagent.
To determine the limiting reagent, we compare the number of moles of reactants using the stoichiometric ratio from the balanced equation.
The stoichiometric ratio between Cr2O3 and Al is 1:2. This means that 1 mole of Cr2O3 reacts with 2 moles of Al.
Using the mole ratio, we can calculate the number of moles of Al required:
Number of moles of Al required = 2 × Number of moles of Cr2O3
= 2 × 0.274 moles
= 0.548 moles

Since the number of moles of Al (0.548 moles) is equal to the number of moles of Al required, it is the limiting reagent. This means that all of the Al will react, and the excess Cr2O3 will be left over.

Step 5: Calculate the moles of chromium produced.
Using the stoichiometric ratio from the balanced equation, we can calculate the number of moles of Cr produced:
Number of moles of Cr = 2 × Number of moles of Al
= 2 × 0.548 moles
= 1.096 moles

Step 6: Calculate the mass of aluminum oxide produced.
Using the molar mass of aluminum oxide (Al2O3), which is 102 g/mol:
Mass of Al2O3 = Number of moles of Al2O3 × Molar mass of Al2O3
= 0.548 moles × 102 g/mol
= 55.896 g

Therefore, the mass of aluminum oxide produced is 55.896 g.

To solve this problem, we need to use stoichiometry, which is the quantitative relationship between the reactants and products in a chemical reaction.

1. Start by writing a balanced chemical equation for the reaction:
2Al + 3Cr2O3 -> 2Al2O3 + 6Cr

The equation shows that for every 3 moles of chromium(III) oxide (Cr2O3), we need 2 moles of aluminum (Al) to produce 2 moles of aluminum oxide (Al2O3) and 6 moles of chromium (Cr).

2. Convert the given masses of chromium(III) oxide and aluminum to moles:
Moles of Cr2O3 = Mass of Cr2O3 / Molar mass of Cr2O3
Moles of Al = Mass of Al / Molar mass of Al

The molar masses are calculated as follows:
Molar mass of Cr2O3 = (2 * Molar mass of Cr) + (3 * Molar mass of O)
Molar mass of Al = Atomic mass of Al

3. Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed. To find the limiting reactant, compare the moles of each reactant using the stoichiometry ratio from the balanced equation.

4. Calculate the moles of chromium produced:
Moles of Cr = Moles of Al * (6/2) (from the stoichiometry ratio)

5. Once you know the moles of chromium produced, calculate the moles of aluminum oxide using the stoichiometry ratio:
Moles of Al2O3 = Moles of Cr * (2/6) (from the stoichiometry ratio)

6. Finally, convert the moles of aluminum oxide to grams:
Mass of Al2O3 = Moles of Al2O3 * Molar mass of Al2O3

By following these steps, you can determine the mass of aluminum oxide produced.