<u>Answer:</u> The theoretical yield of manganese dioxide is 57.42 grams
<u>Explanation:</u>
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)](https://tex.z-dn.net/?f=2Mn%28s%29%2B3O_2%28g%29%5Crightarrow%20MnO_2%28s%29)
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 =
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 =
of manganese dioxide
To calculate the number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
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](https://tex.z-dn.net/?f=0.66mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20manganese%20dioxide%7D%7D%7B87g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20manganese%20dioxide%7D%3D%280.66mol%5Ctimes%2087g%2Fmol%29%3D57.42g)
Hence, the theoretical yield of manganese dioxide is 57.42 grams