Question

Metallic aluminum reacts with MnO2 at elevated temperatures to form manganese metal and aluminum oxide. A mixture of the two reactants is 67.2% mole percent .Find thetheoretical yield (in grams) of manganese from the reaction of 190g of this mixture

Answer 1

4Al + 3 MnO2 =2 Al2O3 +3 Mn mass Al = 190 x 67.2/100=127.68 g moles Al = 127.68 / 26.98 g/mol=4.73 mass MnO2 = 190 - 127.68=62.32 g moles MnO2 = 62.32 / 87g/mol=0.72 ( limiting reactant) moles Mn = 0.72 mass Mn = 0.72x 55 g/mol=39.6 g

Answer 2

Balanced equation:
4Al + 3MnO2 ---> 3Mn + 2Al2O3

The molecular weight of the mixture/reactants:
4Al + 3MnO2 = 4Al + 3 Mn + 6O =

4 Al * 26.9815386 = 107.926
3 Mn * 54.938045 = + 164.814
6 O * 15.9994 = + 95.996
(1mol mixture/368.737g mixture)

From the mol %:
(67.2 mole Al/100 mol mixture)

From the balanced equation:
(4 mol Al/3 mol Mn)

Atomic weight of Mn:
(1mol Mn/54.938g Mn)

Calculation:
Starting from 190 g mixture and using the above factors:

(190 g mixture)
x (1mol mixture/368.737g mixture)
x (67.2 mole Al/100 mol mixture)
x (3 mol Mn/4 mol Al)
x (54.938g Mn/1mol Mn)

With just the numbers:
(190) x (1/368.737) x (67.2/100) x (3/4) x (54.938)

= 14.26725 g Mn

= 14.3 g Mn
(theoretical)