Question

missed this? watch kcv: limiting react, theoretical yield, and percent yield, iwe: 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 b determine the theoretical yield for the reaction. m = g part c determine percent yield for the reaction.

Explanation:

Step 1: Find Moles of Mg

Molar mass of Mg is \( 24.31 \, \text{g/mol} \). Moles of Mg: \( \frac{10.1 \, \text{g}}{24.31 \, \text{g/mol}} \approx 0.4155 \, \text{mol} \).

Step 2: Find Moles of \( O_2 \)

Molar mass of \( O_2 \) is \( 32.00 \, \text{g/mol} \). Moles of \( O_2 \): \( \frac{10.5 \, \text{g}}{32.00 \, \text{g/mol}} \approx 0.3281 \, \text{mol} \).

Step 3: Determine Limiting Reactant

From reaction \( 2\text{Mg} + \text{O}_2
ightarrow 2\text{MgO} \), mole ratio \( \text{Mg}:\text{O}_2 = 2:1 \).
Moles of \( O_2 \) required for Mg: \( \frac{0.4155}{2} = 0.2078 \, \text{mol} \) (less than 0.3281, so Mg is limiting).

Step 4: Theoretical Yield of MgO

Moles of MgO from Mg: \( 0.4155 \, \text{mol} \) (1:1 ratio). Molar mass of MgO: \( 40.31 \, \text{g/mol} \).
Theoretical yield: \( 0.4155 \, \text{mol} \times 40.31 \, \text{g/mol} \approx 16.75 \, \text{g} \).

Step 5: Percent Yield

Actual yield = 12.9 g. Percent yield: \( \frac{12.9}{16.75} \times 100 \approx 77.0\% \).

Answer:

\( 77.0\% \) (or rounded as needed, e.g., 77% or 77.0)