Question

missed this? watch kcv: limiting reactant, theoretical yield, and percent yield we limiting reactant and theoretical yield read section 4.4 you can click on the review link to access the section in your e - text. for the reaction mn(s)+o₂(g)→mno₂(s) compute the theoretical yield of product (in moles) for each of the following initial amounts of reactants part c 28.0 mol mn and 44.8 mol o₂ express your answer using three significant figures.

Explanation:

Step1: Determine the stoichiometric ratio

The balanced equation is $Mn(s)+O_2(g)
ightarrow MnO_2(s)$. The ratio of $Mn$ to $O_2$ to $MnO_2$ is 1:1:1.

Step2: Identify the limiting reactant

The mole - ratio of $Mn$ to $O_2$ is $\frac{n_{Mn}}{n_{O_2}}=\frac{28.0\ mol}{44.8\ mol}=\frac{5}{8}$. Since the stoichiometric ratio of $Mn$ to $O_2$ is 1:1, $Mn$ is the limiting reactant.

Step3: Calculate the theoretical yield

From the stoichiometry, 1 mole of $Mn$ produces 1 mole of $MnO_2$. So, if we have 28.0 mol of $Mn$ (the limiting reactant), the amount of $MnO_2$ produced is also 28.0 mol.

Answer:

28.0 mol