Question

1. aluminum undergoes a synthesis reaction with oxygen to form aluminum oxide:

4 al(s) + 3 o₂(g) → 2 al₂o₃(s)
which of the following are stoichiometric amounts of the two reactants?
quantity of al(s)
2.0 mol
0.60 mol
0.13 mol
108 g
4.0 g
quantity of o₂(g)
1.5 mol
0.45 mol
0.12 mol
96 g
3.0 g

Explanation:

Step1: Analyze the mole ratio

From the balanced equation \(4Al(s)+3O_{2}(g)\to2Al_{2}O_{3}(s)\), the mole ratio of \(Al\) to \(O_{2}\) is \(n(Al):n(O_{2}) = 4:3=\frac{4}{3}\)

Step2: Check each pair of moles

• For \(n(Al)=2.0\space mol\) and \(n(O_{2}) = 1.5\space mol\), \(\frac{n(Al)}{n(O_{2})}=\frac{2.0}{1.5}=\frac{4}{3}\)
• For \(n(Al)=0.60\space mol\) and \(n(O_{2}) = 0.45\space mol\), \(\frac{n(Al)}{n(O_{2})}=\frac{0.60}{0.45}=\frac{4}{3}\)
• For \(n(Al)=0.13\space mol\) and \(n(O_{2}) = 0.12\space mol\), \(\frac{n(Al)}{n(O_{2})}=\frac{0.13}{0.12}

eq\frac{4}{3}\)

Step3: Convert mass to moles

The molar mass of \(Al\) is \(M(Al)=27\space g/mol\), and the molar mass of \(O_{2}\) is \(M(O_{2}) = 32\space g/mol\)

• For \(m(Al)=108\space g\), \(n(Al)=\frac{m(Al)}{M(Al)}=\frac{108}{27}=4\space mol\)

For \(m(O_{2}) = 96\space g\), \(n(O_{2})=\frac{m(O_{2})}{M(O_{2})}=\frac{96}{32}=3\space mol\), \(\frac{n(Al)}{n(O_{2})}=\frac{4}{3}\)

• For \(m(Al)=4.0\space g\), \(n(Al)=\frac{m(Al)}{M(Al)}=\frac{4.0}{27}\space mol\)

For \(m(O_{2}) = 3.0\space g\), \(n(O_{2})=\frac{m(O_{2})}{M(O_{2})}=\frac{3.0}{32}\space mol\), \(\frac{n(Al)}{n(O_{2})}=\frac{4.0/27}{3.0/32}=\frac{4\times32}{27\times3}
eq\frac{4}{3}\)

Answer:

• \(2.0\space mol\) of \(Al(s)\) and \(1.5\space mol\) of \(O_{2}(g)\)
• \(0.60\space mol\) of \(Al(s)\) and \(0.45\space mol\) of \(O_{2}(g)\)
• \(108\space g\) of \(Al(s)\) and \(96\space g\) of \(O_{2}(g)\)