Question

part a
5 mol mn and 5 mol o₂
express your answer using one significant figure.

part b
5 mol mn and 7 mol o₂
express your answer using one significant figure.

Explanation:

1. First, write the balanced chemical - equation for the reaction between Mn and O₂. The reaction is \(2Mn + O_{2}

ightarrow2MnO\).

• From the balanced equation, the mole - ratio of Mn to O₂ is \(n_{Mn}:n_{O_{2}} = 2:1\).

1. Part A:

• Given \(n_{Mn}=5\ mol\) and \(n_{O_{2}} = 5\ mol\).
• Determine the limiting reactant. Calculate the amount of O₂ required to react completely with 5 mol of Mn. According to the mole - ratio, if \(n_{Mn}=5\ mol\), the amount of O₂ required, \(n_{O_{2}\ required}=\frac{1}{2}n_{Mn}\).
• Substitute \(n_{Mn}=5\ mol\) into the formula: \(n_{O_{2}\ required}=\frac{1}{2}\times5\ mol = 2.5\ mol\). Since we have 5 mol of O₂ and only 2.5 mol are required to react with 5 mol of Mn, Mn is the limiting reactant.
• Using the mole - ratio of Mn to MnO (\(n_{Mn}:n_{MnO}=1:1\)), the amount of MnO formed is equal to the amount of the limiting reactant (Mn). So, \(n_{MnO}=5\ mol\).

1. Part B:

• Given \(n_{Mn}=5\ mol\) and \(n_{O_{2}} = 7\ mol\).
• Calculate the amount of O₂ required to react completely with 5 mol of Mn. Using the mole - ratio \(n_{O_{2}\ required}=\frac{1}{2}n_{Mn}\), substitute \(n_{Mn}=5\ mol\) to get \(n_{O_{2}\ required}=\frac{1}{2}\times5\ mol = 2.5\ mol\). Since we have 7 mol of O₂ and only 2.5 mol are required to react with 5 mol of Mn, Mn is the limiting reactant.
• Using the mole - ratio of Mn to MnO (\(n_{Mn}:n_{MnO}=1:1\)), the amount of MnO formed is equal to the amount of the limiting reactant (Mn). So, \(n_{MnO}=5\ mol\).

Step1: Write balanced equation

\(2Mn + O_{2}
ightarrow2MnO\)

Step2: Determine limiting reactant in Part A

Since \(n_{O_{2}\ required}=\frac{1}{2}n_{Mn}=\frac{1}{2}\times5\ mol = 2.5\ mol\) and \(n_{O_{2}\ available}=5\ mol\), Mn is limiting.

Step3: Calculate product amount in Part A

\(n_{MnO}=n_{Mn}=5\ mol\)

Step4: Determine limiting reactant in Part B

Since \(n_{O_{2}\ required}=\frac{1}{2}n_{Mn}=\frac{1}{2}\times5\ mol = 2.5\ mol\) and \(n_{O_{2}\ available}=7\ mol\), Mn is limiting.

Step5: Calculate product amount in Part B

\(n_{MnO}=n_{Mn}=5\ mol\)

Answer:

Part A: 5 mol
Part B: 5 mol