In a reaction, 37.5 g of chromium(III) oxide reacts with 13.3 g of aluminum to produce chromium and aluminum oxide. If 25.7 g of chromium is produced, what mass of aluminum oxide is produced?
This is a regular stoichiometry problem disguised to appear as a limiting reagent problem. Here is a worked example of a stoichiometry problem. Just follow the steps. http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the mass of aluminum oxide produced, we need to start by calculating the moles of chromium(III) oxide and aluminum using their respective masses.
Step 1: Calculate the moles of chromium(III) oxide (Cr2O3).
The molar mass of Cr2O3 = (2 x molar mass of chromium) + (3 x molar mass of oxygen)
The molar mass of chromium (Cr) = 52.0 g/mol
The molar mass of oxygen (O) = 16.0 g/mol
Molar mass of Cr2O3 = (2 x 52.0 g/mol) + (3 x 16.0 g/mol) = 152.0 g/mol
Number of moles of Cr2O3 = mass / molar mass
Number of moles of Cr2O3 = 37.5 g / 152.0 g/mol
Step 2: Calculate the moles of aluminum (Al).
The molar mass of aluminum (Al) = 26.98 g/mol
Number of moles of Al = mass / molar mass
Number of moles of Al = 13.3 g / 26.98 g/mol
Step 3: Determine the limiting reactant.
The limiting reactant is the one with the smallest number of moles. Comparing the number of moles of Cr2O3 and Al, we find that Al has the smallest number of moles.
Step 4: Calculate the moles of chromium produced.
From the balanced chemical equation of the reaction, we know that 1 mole of Cr2O3 produces 2 moles of Cr. Therefore, the number of moles of Cr produced is twice the number of moles of Cr2O3.
Number of moles of Cr = 2 x (Number of moles of Cr2O3)
Step 5: Calculate the mass of aluminum oxide produced.
From the balanced chemical equation, we know that 1 mole of Cr2O3 produces 1 mole of Al2O3. Therefore, the number of moles of Al2O3 produced is the same as the number of moles of Cr2O3.
Number of moles of Al2O3 = Number of moles of Cr2O3
Finally, we can calculate the mass of Al2O3:
Mass of Al2O3 = Number of moles of Al2O3 x molar mass of Al2O3
Note: To calculate the molar mass of Al2O3, we add the molar masses of aluminum (Al) and oxygen (O).
Molar mass of Al2O3 = (2 x molar mass of Al) + (3 x molar mass of O)
Substituting the values and calculating, we can determine the mass of aluminum oxide produced.
To find the mass of aluminum oxide produced, we first need to determine the balanced chemical equation for the reaction.
The balanced equation for the reaction is as follows:
2Cr2O3 + 4Al -> 4Cr + 2Al2O3
From the equation, we can see that for every 2 moles of Cr2O3, we need 4 moles of Al to produce 2 moles of Al2O3.
Now, let's calculate the moles of chromium(III) oxide (Cr2O3) and aluminum (Al) using their given masses.
Moles of Cr2O3 = Mass / Molar mass
Molar mass of Cr2O3 = (2 x Atomic mass of Cr) + (3 x Atomic mass of O)
Atomic mass of Cr = 52 g/mol
Atomic mass of O = 16 g/mol
Moles of Cr2O3 = 37.5 g / [(2 x 52 g/mol) + (3 x 16 g/mol)]
Moles of Cr2O3 ≈ 37.5 g / 152 g/mol
Similarly, calculate the moles of Al:
Moles of Al = 13.3 g / Atomic mass of Al
Atomic mass of Al = 27 g/mol
Moles of Al ≈ 13.3 g / 27 g/mol
Now, we need to find the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed.
To find the limiting reactant, we compare the moles of Cr2O3 and Al. The reactant with fewer moles is the limiting reactant.
From the balanced equation, we know that 2 moles of Cr2O3 reacts with 4 moles of Al to produce 2 moles of Al2O3. Therefore, the ratio of moles of Cr2O3 to Al to Al2O3 is 1:2:1.
Since the moles of Al2O3 formed will be equal to the moles of Cr produced, we need to find the moles of Cr produced.
Moles of Cr = Moles of Cr2O3
Now, we can calculate the moles of Cr produced:
Moles of Cr = Moles of Cr2O3 ≈ 37.5 g / 152 g/mol (as calculated earlier)
Now, calculate the mass of Al2O3 (aluminum oxide) produced.
Moles of Al2O3 = Moles of Cr ≈ Moles of Cr2O3 ≈ 37.5 g / 152 g/mol
Mass of Al2O3 = Moles of Al2O3 x Molar mass of Al2O3
Molar mass of Al2O3 = (2 x Atomic mass of Al) + (3 x Atomic mass of O)
Atomic mass of O = 16 g/mol
Mass of Al2O3 = (37.5 g / 152 g/mol) x [(2 x 27 g/mol) + (3 x 16 g/mol)]
Finally, calculate the mass of Al2O3 to find the answer.