Question

22. Consider the reaction: solid aluminum and oxygen gas react to form solid aluminum oxide. a. Write and balance the equation. b. How many moles of aluminum are required to react with 47.2 grams of oxygen? c. If 14.0 g of aluminum reacts with oxygen gas, how many grams of aluminum oxide will form?

Answer 1

22. a. 4 Al_(s) + 3 O_2(g) $\rightarrow$ 2 Al₂O_3(s)

b. Moles of oxygen = mass / molar mass = 47.2 / 32 = 1.475 moles

From the equation: moles of al = 4 / 3 * moles of O₂ = 4 / 3 * 1.475 = 1.967 moles

c. Moles of Al = mass / molar mass = 14 / 27 = 0.52 moles

From the equation : moles of Al₂O₃ = 1 / 2 * moles of Al = 1/ 2 * 0.52 = 0.26 moles

Mass of Al₂O₃ = moles * molar mass = 0.26 * 102 = 26.52 g

d. Atoms of Al = Avogadr number * number of moles = 6.022 * 10²³ * 0.52 = 3.13 * 10²³ atoms

e. % yield = mass formed / theoretical mass fromed * 100 = 19.7 / 26.52 * 100 = 74.28%