Question

When 110g of Al and 200g of MnO are heated to initiate following reaction Al + MnO rightarrow Al_2O_3 + Mn How much Mn was produced and which reactant has starting material left and how much?

Answer 1

From balanced equation:

      6Al             +          9MnO   --------->   3Al2O3        +    9Mn

   6 x (27) g                 9 X (70.9) g 3 X (102) g              9 X (54.9) g

From this equation

162 g of Al react with 638.1 g of MnO to produce 306 g of Al2O3 and 494.1 g of Mn

Given that 110 g of Al and 200 g of MnO react with each other. Here the limiting reagent can be calculated from above relation in grams as:

The MnO required for comlete consumption of 110 g of Al = 110 X 638.1 / 162

= 70191 / 162 = 433.3 g of MnO

But given amount of MnO = 200 g

So, the limiting reagent here is MnO that is = 200g

Because, this amount of MnO will react with only 50.8 g of Al leaving unreacted Al = 110 - 50.8 = 59.2 g (left as starting materiel)

Calculation: Amont of Al used for reaction with 200 g of MnO

= 200 X 162 / 638.1

= 50.8 g of Al would be consumed.

Now, Produced Mn after reaction of 200 g of MnO and 50.8 g of Al:

= 200 X 494.1 / 638.1

= 98820 / 638.1 = 154.9 g of Mn