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:
2mg(s)+o2(g)→2mgo(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 b
determine the theoretical yield for the reaction.
part c
determine percent yield for the reaction.

Explanation:

Part B: Determine the theoretical yield for the reaction

Step1: Calculate moles of Mg

Molar mass of \(Mg = 24.305\ g/mol\).
Number of moles of \(Mg,n_{Mg}=\frac{m_{Mg}}{M_{Mg}}=\frac{10.1\ g}{24.305\ g/mol}\approx0.4156\ mol\)

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

Molar mass of \(O_{2}=32\ g/mol\).
Number of moles of \(O_{2},n_{O_{2}}=\frac{m_{O_{2}}}{M_{O_{2}}}=\frac{10.5\ g}{32\ g/mol}\approx0.3281\ mol\)

Step3: Use stoichiometry from the balanced equation \(2Mg(s)+O_{2}(g)\to2MgO(s)\)

From the equation, the mole ratio of \(Mg\) to \(MgO\) is \(2:2 = 1:1\), and mole ratio of \(O_{2}\) to \(MgO\) is \(1:2\)
If \(Mg\) is the limiting reactant, moles of \(MgO\) produced \(n_{MgO}=n_{Mg}\approx0.4156\ mol\)
If \(O_{2}\) is the limiting reactant, moles of \(MgO\) produced \(n_{MgO}=2\times n_{O_{2}}=2\times0.3281\ mol = 0.6562\ mol\)
Since \(0.4156\ mol<0.6562\ mol\), \(Mg\) is the limiting reactant.

Step4: Calculate mass of \(MgO\) (theoretical yield)

Molar mass of \(MgO = 40.3044\ g/mol\)
Theoretical yield \(m_{theoretical}=n_{MgO}\times M_{MgO}=0.4156\ mol\times40.3044\ g/mol\approx16.75\ g\)

Part C: Determine percent yield for the reaction

Step1: Use the percent - yield formula

Percent yield formula is \(\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%\)
Given actual yield \(m_{actual}=11.3\ g\) and theoretical yield \(m_{theoretical}\approx16.75\ g\)
\(\text{Percent Yield}=\frac{11.3\ g}{16.75\ g}\times100\% \approx 67.46\%\)

Answer:

Part B: \(16.75\ g\)
Part C: \(67.5\%\)