Question

when aluminum oxidizes in air, it forms aluminum oxide (al₂o₃):
4al (s) + 3o₂ (g) —→ 2al₂o₃ (s)
a 51 g sheet of aluminum oxide formed completely in excess oxygen. how many grams of aluminum were oxidized?
○ 27 g
○ 51 g
○ 102 g
○ 54 g
question 12
1 pts
consider this hypothetical reaction: a + 3b —→ c + 3d. assume that the equation is balanced. in which of the reactant mole ratios given will reactant b be the limiting reagent?
○ a/b
○ a/5b
○ a/4b
○ 2a/7b

Explanation:

Step1: Calculate moles of \(Al_2O_3\)

The molar mass of \(Al_2O_3\) is \(M = 2\times27+3\times16=102\space g/mol\).
The number of moles of \(Al_2O_3\), \(n=\frac{m}{M}=\frac{51\space g}{102\space g/mol} = 0.5\space mol\)

Step2: Use mole ratio from the balanced equation

From the balanced equation \(4Al(s)+3O_2(g)\to2Al_2O_3(s)\), the mole ratio of \(Al\) to \(Al_2O_3\) is \(4:2 = 2:1\).
If \(n(Al_2O_3)=0.5\space mol\), then \(n(Al)=2\times n(Al_2O_3)=2\times0.5\space mol = 1\space mol\)

Step3: Calculate mass of \(Al\)

The molar mass of \(Al\) is \(M = 27\space g/mol\).
The mass of \(Al\), \(m=n\times M=1\space mol\times27\space g/mol = 27\space g\)

Answer:

27 g