Question

The reaction 2 AL+ 3M nO Al 2O3
+ 3 Mn proceeds till the limiting
substance is consumed. A mixture
of 220g
of Al and 400g of
MnO was heated to initiate
the reaction which initial substance
remained in excess and by how
much?​

Answer 1

Answer:

2Al + 3MnO Al203+ 3Mn

initial: 2 moles of Al and 3moles of MnO

final: given 220 g of Al so

no of moles= 220/26 = 8.46

then 400 g of MnO so

no of moles = 400/52= 7.40

so therefore final moles = approximately 8.46 and 7.40 respectively

then Al = 8.46/2 = 4.23

and MnO =7.40/3= 2.46

this shows Al is excess and by 4.23-2.46= 1.77 moles