Calculate the theoretical yield of product (in moles) for each initial amount of reactants.
3 Mn(s) + 2 O₂(g) → Mn₃O₄(s)
a. 3 mol Mn, 2 mol O₂
b. 4 mol Mn, 7 mol O₂
c. 27.5 mol Mn, 43.8 mol O₂

Question

Calculate the theoretical yield of product (in moles) for each initial amount of reactants.

3 Mn(s) + 2 O₂(g) → Mn₃O₄(s)

a. 3 mol Mn, 2 mol O₂

b. 4 mol Mn, 7 mol O₂

c. 27.5 mol Mn, 43.8 mol O₂

Accepted Answer

Hello, everyone. Today, we have the following problem, calculate the theoretically yield of the product in moles for each initial amount of reactants. So we have to use the mole ratios to convert the reactants into the product. And so we are given our four moles of our manganese and seven moles of oxygen. So for our manganese, we take our four moles of it and we have to multiply by the multiple ratio. That is saying that if we look at our equation, we see how many moles of manganese we have, which is three, we put that in the denominator so that we can cancel it and it's out. And then on our enumerator, we put the moles of our manganese oxide that we formed, which is just one. And we see now when our units cancel L we write an answer of 1.333 moles of that manganese oxide. We do the same for our oxygen. So we have our seven moles of our oxygen gas. And we do the same thing with our multiple conversion. We say that we have one mole of our manganese sign in ratio with our two moles of oxygen gas, our units will cancel out and this will equal 3.5 moles of our manganese oxide. Since manganese produce the less or not, it is the limiting reactant. Well, oxygen gas of the excess and the product produced when the limiting reactant fully reacted is the theoretical yield. And so essentially, when we use this 1.33 moles of our manganese oxide, we yielded one mole of our manganese oxide or our product making your answer and your choice. A and with that, we have solved the problem overall, I hope is help. And until next time.

Calculate the theoretical yield of product (in moles) for each initial amount of reactants.
3 Mn(s) + 2 O₂(g) → Mn₃O₄(s)
a. 3 mol Mn, 2 mol O₂
b. 4 mol Mn, 7 mol O₂
c. 27.5 mol Mn, 43.8 mol O₂

Verified Solution

Key Concepts

Stoichiometry

Limiting Reactant

Theoretical Yield

Calculate the theoretical yield of product (in moles) for each initial amount of reactants. 3 Mn(s) + 2 O2(g) → Mn3O4(s) a. 3 mol Mn, 2 mol O2b. 4 mol Mn, 7 mol O2c. 27.5 mol Mn, 43.8 mol O2

Verified Solution

Key Concepts

Stoichiometry

Limiting Reactant

Theoretical Yield

Calculate the theoretical yield of product (in moles) for each initial amount of reactants.
3 Mn(s) + 2 O₂(g) → Mn₃O₄(s)
a. 3 mol Mn, 2 mol O₂
b. 4 mol Mn, 7 mol O₂
c. 27.5 mol Mn, 43.8 mol O₂