Question

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1. consider the reaction ( 2 mathrm{al}(mathrm{s})+mathrm{fe}_{2} mathrm{o}_{3}(mathrm{~s}) \to 2 mathrm{fe}(mathrm{l})+mathrm{al}_{2} mathrm{o}_{3}(mathrm{~s}) ). if ( 60.0 mathrm{~g} ) of ( mathrm{al} ) is reacted with excess ( mathrm{fe}_{2} mathrm{o}_{3} ), determine the amount (in moles) and mass (in grams) of ( mathrm{al}_{2} mathrm{o}_{3} ) produced. ( left(m_{mathrm{al}_{2} mathrm{o}_{3}}=101.96 mathrm{~g} / mathrm{mol}

ight) )

Explanation:

Step1: Calculate the moles of Al

The molar mass of Al (\(M_{Al}\)) is \(26.98\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 60.0\space g\) and \(M = 26.98\space g/mol\)
\(n_{Al}=\frac{60.0\space g}{26.98\space g/mol}\approx2.22\space mol\)

Step2: Determine the moles of \(Al_{2}O_{3}\) using stoichiometry

From the balanced equation \(2Al(s)+Fe_{2}O_{3}(s)\to2Fe(l)+Al_{2}O_{3}(s)\), the mole ratio of \(Al\) to \(Al_{2}O_{3}\) is \(2:1\)
So \(n_{Al_{2}O_{3}}=\frac{1}{2}n_{Al}\)
Substituting \(n_{Al} = 2.22\space mol\), we get \(n_{Al_{2}O_{3}}=\frac{1}{2}\times2.22\space mol = 1.11\space mol\)

Step3: Calculate the mass of \(Al_{2}O_{3}\)

Using the formula \(m = nM\), where \(n = 1.11\space mol\) and \(M = 101.96\space g/mol\)
\(m_{Al_{2}O_{3}}=1.11\space mol\times101.96\space g/mol\approx113\space g\)

Answer:

The amount of \(Al_{2}O_{3}\) is \(1.11\space mol\) and the mass is \(113\space g\)