Question

Considering the limiting reactant, what is the mass of manganese produced from 25.0 g of manganese(iv) oxide (86.94 g/mol) and 25.0 g of aluminum metal? 3 mno2(l) + 4 al(l) 3 mn(l) + 2 al2o3(s)

Answer 1

Answer: 15.749g of manganese will be produced from the given reaction.

Explanation: We are given a chemical reaction:

$3MnO_2(l)+4Al(l)\rightarrow 3Mn(l)+2Al_2O_3(s)$

To know which of the reactant is the limiting reagent, we will have to first find out the moles of each reactant. Formula used for calculating the moles is:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Moles of Manganese(IV) oxide:

Given mass = 25 g

Molar mass = 86.94 g/mol

$Moles=\frac{25g}{86.94g/mol}=0.2875moles$

Moles of aluminium:

Given mass = 25 g

Molar mass = 26.982 g/mol

$Moles=\frac{25g}{26.982g/mol}=0.9265moles$

By stoichiometry of the reaction:

3 moles of manganese (IV) oxide reacts with 4 moles of aluminium.

So, 0.2875 moles of manganese (IV) oxide will react with = $\frac{4}{3}\times 0.2875$ = 0.3833 moles of aluminium.

As, the moles of aluminium is more than the required. Hence, it is the excess reagent. Therefore, manganese (IV) oxide is the limiting reagent because it limits the formation of the product.

By Stoichiometry of the reaction:

3 moles of manganese (IV) oxide produces 3 moles of manganese.

So, 0.2875 moles of manganese (IV) oxide will produce = $\frac{3}{3}\times 0.2875$ = 0.2875 moles of manganese.

To calculate the mass of manganese, we use equation 1:

Molar mass of manganese = 54.938 g/mol

Putting values in equation 1, we get:

$0.2875mol=\frac{\text{Mass of manganese}}{54.938g/mol}$

Mass of manganese = 15.794g