The balanced reaction is:
4Fe + 3O2 --> 2Fe2O3
Stoichiometrically:
(1.0 mol Fe)(2 mol Fe2O3 / 4 mol Fe) = 0.50 mol Fe2O3
If the actual yield is only 0.325 mol Fe2O3, the % yield can be calculated by dividing actual by theoretical yield:
0.325 / 0.5 x 100% = 65% yield
<span>% by mass = mass solute x 100 / mass solution
45.5 % = mass solute NaF x 100 / 34.2
mass solute NaF = 34.2 x 45.5 /100=15.6 g
molae solute NaF = 15.6 g ( 1 mol / 41.9887 g)= 0.372</span>
C not sure 100 percent but my best guess
Answer:
Yes.
Explanation:
It will react. Because they are compounds
A) Na2S
b) AlF3
c) O2
d) C6H12O6
