Aurous is a cation of gold. Gold takes the name "aurum" (Au) with atomic number of 79. In its purest form, the element is bright, slightly yellow, soft, ductile, and malleable. The charge of aurous is +1. Sulfide, on the other hand, has a charge of -2.
Hence, the chemical formula of the compound is Au₂S and its systematic name is gold (I) sulfide.
<span>4FeS2 + 11O2 = 2Fe2O3 + 8SO2</span>
Percent yield is calculated as the actual yield divided by the theoretical yield multiplied by 100.
Actual yield = 55 g ( 1 mol / 159.69 g ) = 0.34 mol Fe2O3
To find for the theoretical yield, we first determine the limiting reactant.
100 g O2 ( 1 mol / 32 g) = 3.13 mol O2
200 g FeS2 (1 mol / 119.98g) = 1.67 mol FeS2
Therefore, the limiting reactant is O2.
Theoretical yield = 3.13 mol O2 ( 2 mol Fe2O3 / 11 mol O2 ) = 0.57 mol Fe2O3
Percent yield = (0.34 mol / 0.57 mol) x 100 = 59.74%
Answer:
I don't know
Explanation:
Maybe they shouldn't copy each other
n=20 mol
(NH)4 SO4
Atomic masses :
N- 14
H- 1
S- 32
O- 16
Therefore M= 14×2 + 1×8 + 32 + 16×4
= 132
m= nM
= 20×132
= 2640g
