Question

Considering the reaction 2Al₂O₃+3C ----> 4Al +3CO₂

A- How many moles of C are needed to react with 7.75 moles of Al₂O₃?

B- How many grams of Al result when 355 grams of Al₂O₃ react?

Answer 1

Reaction : 2 Al2O3 + 3C -------> 4Al + 3CO2

Here We know that,

In terms of moles       2 moles of Al2o3 reacts with 3moles of C TO Form 4moles of Al & 3moles of CO2

A .    2moles of Al2O3 --------> 3moles of C

       7.75moles of Al2O3 -------- ? moles of C

Moles of C = 3 * 7.75 / 2

                  =   11.63 Moles

11.63moles of C Needed to react with 7.75moles of Al2O3

B . 2 moles of Al2O3 reacts with 3moles of C To form " 4moles of Al "

      2moles of Al2O3 --------> 4moles Al

     355 grams of Al2O3 -------- ? grams of Al

We know that , Mole = Wt / Mwt

2 ( Wt / Mwt of Al2O3) = 4 (Wt / Mwt of Al)

        2 ( 355 / 101.8 ) = 4 ( Wt / 26.9)

                  6.97         = 4(wt/26.9)

                         Wt      = 6.97 * 26.9 / 4

                       Wt         = 46.87 grams of Al

         355grams of Al2O3 React to form 46.87 grams of Al