Use the ideal gas law:
PV = nRT
so, T = PV / nR
n=0.5
V= 120 dm^3 = 120 L (1 dm^3 = 1 L)
R = 1/12
P = 15,000 Pa = 0.147 atm (1 pa = 9.86 10^{-6} )
Put the values:
T = PV / nR
T = (0.147) (120) / (0.5) (1/12)
T= 426 K
10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
What ois Theoretical yield ?
The quantity of a product obtained from a reaction is expressed in terms of the yield of the reaction.
The amount of product predicted by stoichiometry is called the theoretical yield, whereas the amount obtained actually is called the actual yield.
- As 2 moles (111.68 g) of Fe consumes 213 gm of Cl₂ to produce 2FeCl₃
Therefore ,
10 gm of Fe will consumes = 213 / 111.68 x 10 = 19 gm Cl₂
- As 2 moles (111.68 g) of Fe produces 2 mole (324 gm) of FeCl₃
Therefore ,
10 gm of Fe will produces = 324 / 111.68 x 10 = 29 gm FeCl₃
Hence , 10 gm of Fe will consumes 19 gm Cl₂ and will produces 29 gm FeCl₃.
Learn more about Theoretical yield here ;
brainly.com/question/12704041
#SPJ1
Answer:
hello I am interested in the position and would like to know if you have any questions please feel free to contact me at any time if
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.
