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:
[tex]2Mn(s)+3O_2(g)\rightarrow MnO_2(s)[/tex]
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 = [tex]\frac{2}{3}\times 2=1.33mol[/tex] 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 = [tex]\frac{1}{3}\times 2=0.66mol[/tex] of manganese dioxide
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of manganese dioxide = 87 g/mol
Moles of manganese dioxide = 0.66 moles
Putting values in above equation, we get:
[tex]0.66mol=\frac{\text{Mass of manganese dioxide}}{87g/mol}\\\\\text{Mass of manganese dioxide}=(0.66mol\times 87g/mol)=57.42g[/tex]
Hence, the theoretical yield of manganese dioxide is 57.42 grams
