Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield, we: finding limiting reactant and theoretical yield. read section 8.6. you can click on the review link to access the section in your etext. 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, 12.9 g of mgo are collected. part a determine the limiting reactant for the reaction. express your answer as a chemical formula. a chemical reaction does not occur for this question. submit request answer part b

Explanation:

Step1: Calculate moles of Mg

Molar mass of Mg is \(24.31\space g/mol\). Moles of Mg = \(\frac{10.1\space g}{24.31\space g/mol} \approx 0.415\space mol\).

Step2: Calculate moles of \(O_2\)

Molar mass of \(O_2\) is \(32.00\space g/mol\). Moles of \(O_2\) = \(\frac{10.5\space g}{32.00\space g/mol} \approx 0.328\space mol\).

Step3: Determine mole ratio from reaction

The reaction is \(2Mg(s) + O_2(g)
ightarrow 2MgO(s)\). The mole ratio of \(Mg\) to \(O_2\) is \(2:1\).

Step4: Find required moles of \(O_2\) for Mg

For \(0.415\space mol\) of Mg, moles of \(O_2\) required = \(\frac{0.415\space mol}{2} = 0.2075\space mol\). But we have \(0.328\space mol\) of \(O_2\), which is more than required. Alternatively, moles of Mg required for \(0.328\space mol\) of \(O_2\) is \(2\times0.328 = 0.656\space mol\), but we have only \(0.415\space mol\) of Mg. So Mg is the limiting reactant.

Answer:

Mg