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Consider the reaction below. 2Al_(2)O_(3)arrow 4Al+3O_(2) How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed? 17.7 mol 19.9 mol 39.8 mol 53.0 mol

Question

Consider the reaction below.
2Al_(2)O_(3)arrow 4Al+3O_(2)
How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed?
17.7 mol
19.9 mol
39.8 mol
53.0 mol

Consider the reaction below. 2Al_(2)O_(3)arrow 4Al+3O_(2) How many moles of oxygen are produced when 26.5 mol of aluminum oxide are decomposed? 17.7 mol 19.9 mol 39.8 mol 53.0 mol

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SerenVeteran · Tutor for 11 years

Answer

\(39.8 \mathrm{~mol}\)

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1. From the balanced chemical equation \(2 \mathrm{Al}_{2} \mathrm{O}_{3} \rightarrow 4 \mathrm{Al} + 3 \mathrm{O}_{2}\), we can determine that 2 moles of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) produce 3 moles of \(\mathrm{O}_{2}\).2. To find out how many moles of \(\mathrm{O}_{2}\) are produced from 26.5 moles of \(\mathrm{Al}_{2} \mathrm{O}_{3}\), we set up a proportion based on the stoichiometry of the reaction:\[\frac{2 \text{ moles } \mathrm{Al}_{2} \mathrm{O}_{3}}{3 \text{ moles } \mathrm{O}_{2}} = \frac{26.5 \text{ moles } \mathrm{Al}_{2} \mathrm{O}_{3}}{x \text{ moles } \mathrm{O}_{2}}\]3. Solving for \(x\) gives us:\[x = \frac{26.5 \text{ moles } \mathrm{Al}_{2} \mathrm{O}_{3} \times 3 \text{ moles } \mathrm{O}_{2}}{2 \text{ moles } \mathrm{Al}_{2} \mathrm{O}_{3}}\]4. Calculating the value of \(x\):\[x = \frac{26.5 \times 3}{2} = \frac{79.5}{2} = 39.75 \text{ moles } \mathrm{O}_{2}\]5. The closest option to this value is 39.8 mol.
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