Question

for the reaction shown compute the theoretical yield of product in moles each of the initial quantities of reactants. 2 Mn(s)+3 O2 (g) _____ MnO2(s) 2 mol Mn , 2 mol O2

Answer 1

Answer: The theoretical yield of manganese dioxide is 57.42 grams

Explanation:

We are given:

Moles of manganese = 2 moles

Moles of oxygen gas = 2 moles

For the given chemical equation:

$2Mn(s)+3O_2(g)\rightarrow MnO_2(s)$

By Stoichiometry of the reaction:

3 moles of oxygen gas reacts with 2 moles of manganese

So, 2 moles of oxygen gas will react with = $\frac{2}{3}\times 2=1.33mol$ of manganese

As, given amount of manganese is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of oxygen gas produces 1 mole of manganese dioxide

So, 2 moles of oxygen gas will produce = $\frac{1}{3}\times 2=0.66mol$ of manganese dioxide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of manganese dioxide = 87 g/mol

Moles of manganese dioxide = 0.66 moles

Putting values in above equation, we get:

$0.66mol=\frac{\text{Mass of manganese dioxide}}{87g/mol}\\\\\text{Mass of manganese dioxide}=(0.66mol\times 87g/mol)=57.42g$

Hence, the theoretical yield of manganese dioxide is 57.42 grams

Answer 2

Answer:

2 mole MnO₂

Explanation:

2Mn(s) + 2O₂(g) => 2MnO₂(s)