In a reaction, 34.0 g of chromium(III) oxide reacts with 12.1 g of aluminum to produce chromium and aluminum oxide. If 23.3 g of chromium is produced, what mass of aluminum oxide is produced?

Short Answer

Expert verified
The mass of aluminum oxide produced in the reaction is approximately \(22.85\ g\).

Step by step solution

01

Find the molar masses of the involved compounds

First, we need to find the molar masses of Cr2O3, Al, Cr, and Al2O3. To do this, we will use their atomic masses from the periodic table: - Atomic mass of Cr: 51.996 g/mol - Atomic mass of Al: 26.982 g/mol - Atomic mass of O: 16.00 g/mol Now, we can find the molar masses of chromium(III) oxide and aluminum oxide: - Molar mass of Cr2O3: \(2(51.996) + 3(16.00) = 151.99\) g/mol - Molar mass of Al2O3: \(2(26.982) + 3(16.00) = 101.96\) g/mol
02

Convert the given masses to moles

Next, we will convert the given masses of chromium(III) oxide, aluminum, and chromium to moles using their molar masses: - Moles of Cr2O3: \(\frac{34.0\ g}{151.99\ \text{g/mol}} = 0.2238\) mol - Moles of Al: \(\frac{12.1\ g}{26.982\ \text{g/mol}} = 0.4485\) mol - Moles of Cr: \(\frac{23.3\ g}{51.996\ \text{g/mol}} = 0.4477\) mol
03

Use stoichiometry to determine moles of aluminum oxide produced

The balanced chemical reaction between chromium(III) oxide and aluminum is: \(Cr_2O_3 + 2Al \rightarrow 2Cr + Al_2O_3\) Using stoichiometry, for 2 moles of aluminum consumed, 1 mole of aluminum oxide is produced. We have already converted the mass of aluminum to moles, so we can use that to find the moles of aluminum oxide produced: Moles of Al2O3: \(\frac{1\ mol\ Al_2O_3}{2\ mol\ Al} × 0.4485\ mol\ Al = 0.2243 \ mol\ Al_2O_3\)
04

Convert moles of aluminum oxide to mass

Finally, we can convert the moles of aluminum oxide back to its mass using its molar mass: Mass of Al2O3: \(0.2243\ mol\ Al_2O_3 × 101.96\ \text{g/mol} = 22.85\ g\) So, the mass of aluminum oxide produced is approximately \(22.85\ g\).

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