Chapter 2: Problem 98

In a reaction, \(34.0 \mathrm{~g}\) of chromium(III) oxide reacts with \(12.1 \mathrm{~g}\) of aluminum to produce chromium and aluminum oxide. If \(23.3 \mathrm{~g}\) 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 \(34.2 \mathrm{~g}\).

Step by step solution

Write the balanced chemical equation for the reaction.

The given reaction is between chromium(III) oxide (Cr2O3) and aluminum (Al) to form chromium (Cr) and aluminum oxide (Al2O3). The balanced chemical equation for the given reaction is: \(2\,Cr_{2}O_{3} + 3\,Al \rightarrow 4\,Cr + 3\,Al_{2}O_{3}\)

Calculate the molar mass of each reactant and product.

The next step is to calculate the molar mass of each substance involved in the reaction. Use the periodic table to find the atomic mass of each element. Molar mass of \(Cr_2O_3\): (2 x 51.996) + (3 x 16.000) = 103.992 + 48.000 = 151.992 g/mol Molar mass of \(Al\): 26.981 g/mol Molar mass of \(Cr\): 51.996 g/mol Molar mass of \(Al_2O_3\): (2 x 26.981) + (3 x 16.000) = 53.962 + 48.000 = 101.961 g/mol

Convert grams of given substances to moles.

Now, we need to convert the given mass of chromium oxide, aluminum, and chromium produced into moles. Moles of \(Cr_2O_3\) = (34.0 g) / (151.992 g/mol) = 0.2237 mol Moles of \(Al\) = (12.1 g) / (26.981 g/mol) = 0.4486 mol Moles of \(Cr\) = (23.3 g) / (51.996 g/mol) = 0.4481 mol

Identify limiting reactant and calculate the moles of aluminum oxide formed.

To find the limiting reactant, we can compare the mole ratios: Mole ratio of chromium oxide to aluminum = [Moles of \(Cr_2O_3\)] / 2 : [Moles of \(Al\)] / 3 0.2237 / 2 : 0.4486 / 3 0.1119 : 0.1495 Since 0.1119 < 0.1495, chromium oxide is the limiting reactant. Now, we can use the stoichiometry from the balanced chemical equation to find the moles of aluminum oxide formed: Moles of \(Al_2O_3\) formed = (3/2) * Moles of \(Cr_2O_3\) = (3/2) * 0.2237 = 0.3356 mol

Convert moles of aluminum oxide to grams and find the mass of aluminum oxide produced.

Finally, use the molar mass of aluminum oxide to convert the moles of aluminum oxide formed to grams: Mass of \(Al_2O_3\) = (0.3356 mol) * (101.961 g/mol) = 34.2 g So, the mass of aluminum oxide produced in the reaction is approximately 34.2 g.

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

Short Answer

Step by step solution

Write the balanced chemical equation for the reaction.

Calculate the molar mass of each reactant and product.

Convert grams of given substances to moles.

Identify limiting reactant and calculate the moles of aluminum oxide formed.

Convert moles of aluminum oxide to grams and find the mass of aluminum oxide produced.

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