Manganese metal can be prepared by the thermite process: 4Al(s) . 3MnO2(s). -->. 3Mn(l). 2Al2O3(s) molar masses, in g/mol: 26.98. 86.94 54.94 101.96 If 203 g of Al and 472 g of MnO2 are mixed, which is the limiting reactant?

MnO2 is the limiting reactant

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

4Al(s) + 3MnO2(s) —> 3Mn(l) + 2Al2O3(s)

Step 2:

Determination of the masses of Al and MnO2 that reacted from the balanced equation. This is illustrated below:

Molar Mass of Al = 26.98g/mol

Mass of Al from the balanced equation = 4 x 26.98 = 107.92g

Molar Mass of MnO2 = 86.94g/mol

Mass of MnO2 from the balanced equation = 3 x 86.94 = 260.82g

From the balanced equation above,

107.92g of Al reacted with 260.82g of MnO2 reacted.

Step 3:

Determination of the limiting reactant. This is illustrated below:

Let us consider using all the 203g of Al given to verify if there will be any leftover for MnO2. This is illustrated below:

From the balanced equation above,

107.92g of Al reacted with 260.82g of MnO2,

Therefore, 203g of Al will react with = (203 x 260.82)/107.92 = 490.61g of MnO2.

We can see that the mass of MnO2 that will react with 203g of Al is 490.61g which far greater than the 472g that was given. Therefore, it is not acceptable.

Now let us consider using all 472g of MnO2 given to verify if there will be any leftover for Al. This is shown below:

From the balanced equation above,

107.92g of Al reacted with 260.82g of MnO2,

Therefore, Xg of Al will react with 472g of MnO2 i.e

Xg of Al = (107.92 x 472)/260.82

Xg of Al = 195.30g

From the calculations above, we can see clearly that there are leftover for Al as 195.30g out of 203g reacted.

Therefore, MnO2 is the limiting reactant

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