Question

2.0 g sample of mixture of Si0, and Fe2O3 on very strong heating leaves a residue weighing 1.96 g. The reaction
responsible for loss of weight is
Fe2O3(s) Fe3O4(s) + 0,(9)
What is the percentage by mass of S10, in original sample? (Atomic weight of Fe=56)
A) 10%
B) 20%
C) 40%
D) 60%​

Answer 1

Answer:

If we make the equation from the given question then we will get the equation as: SiO2 + 3Fe3O4 -> 2Fe3O4 + SiO2+1/2O2.

As from the question we know that the reactant will have total mass of 2g and the Fe3O4 + SiO2 is having weight 1.96g and thus the weight of the oxygen will be 0.02g.

3 moles of Fe3O4 will give 1/2O2.

Thus, 1 O2 will give 3*160/16( molar mass of Fe3O4=160g and O2 is 32g).

So, 1 O2 will be 30 moles so 0.04 will give 30*0.04 = 1.2g.

So, the mass of SiO2 will be (2-1.2)g = 0.8g.

Mass percentage of SiO2 will be 0.8/2*100 = 40%.