<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%
The correct answer is A) the number of electrons that fill the outer shell.
Brady
Answer:
The definition of atom: The Atom is the basic unit of matter.
Types of atoms: There are 3 types of atoms, protons, neutrons, and electrons.
Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
