Question

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magnesium oxide can be made by heating magnesium metal in the presence of oxygen
the balanced equation for the reaction is:
2 mg(s) + o₂(g) → 2 mgo(s)
when 10.1 g of mg are allowed to react with 10.5 g of o₂, 11.3 g of mgo are collected.
part a
determine the limiting reactant for the reaction.
express your answer as a chemical formula.

Explanation:

Step1: Calculate the number of moles of \(Mg\)

The molar mass of \(Mg\) is \(M_{Mg}=24.31\space g/mol\). The number of moles of \(Mg\), \(n_{Mg}=\frac{m_{Mg}}{M_{Mg}}\), where \(m_{Mg} = 10\space g\). So \(n_{Mg}=\frac{10\space g}{24.31\space g/mol}\approx0.411\space mol\)

Step2: Calculate the number of moles of \(O_{2}\)

The molar mass of \(O_{2}\) is \(M_{O_{2}} = 32\space g/mol\). The number of moles of \(O_{2}\), \(n_{O_{2}}=\frac{m_{O_{2}}}{M_{O_{2}}}\), where \(m_{O_{2}}=10.5\space g\). So \(n_{O_{2}}=\frac{10.5\space g}{32\space g/mol}\approx0.328\space mol\)

Step3: Use the stoichiometry of the reaction

From the balanced equation \(2Mg(s)+O_{2}(g)\to2MgO(s)\), the mole ratio of \(Mg\) to \(O_{2}\) is \(2:1\).
For \(n_{Mg} = 0.411\space mol\), the moles of \(O_{2}\) required for complete reaction of \(Mg\) is \(n_{O_{2}\text{(required)}}=\frac{n_{Mg}}{2}=\frac{0.411\space mol}{2}=0.206\space mol\)
Since \(n_{O_{2}\text{(available)}} = 0.328\space mol>n_{O_{2}\text{(required)}}=0.206\space mol\)

For \(n_{O_{2}} = 0.328\space mol\), the moles of \(Mg\) required for complete reaction of \(O_{2}\) is \(n_{Mg\text{(required)}}=2n_{O_{2}}=2\times0.328\space mol = 0.656\space mol\)
Since \(n_{Mg\text{(available)}}=0.411\space mol < n_{Mg\text{(required)}} = 0.656\space mol\)

Answer:

\(Mg\)