.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.
Answer:
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
Explanation:
The equilibrium constant will remain the same irrespective of the amount of reactant taken.
Let us calculate the equilibrium constant of the reaction.
Kc=![\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPCl_%7B3%7D%5D%5BCl_%7B2%7D%5D%7D%7B%5BPCl_%7B5%7D%5D%7D)
Let us calculate the moles of each present at equilibrium

molar mass of PCl₅=208
molar mass of PCl₃=137
molar mass of Cl₂=71
moles of PCl₅ = 
moles of PCl₃= 
moles of Cl₂ = 
the volume is 5 L
So concentration will be moles per unit volume
Putting values
Kc = 
Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

Initial 0.02 0.06 0.04
Change -x +x +x
Equilibrium 0.02-x 0.06+x 0.04+x
Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2
Putting values
0.024 = 
Solving



x = -0.01
so the new moles of
PCl₅ = 0.02 + 0.01 =0.03
PCl₃ = 0.06-0.01 = 0.05
Cl₂ = 0.04-0.01 = 0.03
mass of each will be:
mass= moles X molar mass
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
Answer:
a) C3H8 + 502 ----> 3CO2 + 4H20.
combustion reaction
b) HBr + KOH ----> KBr + H2O
Neutralization reaction
c)AlBr3 + 3 KCl----> 3 KBr + AlCl3
Redox reaction
Answer:
ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)
Explanation:
First, we will write the molecular equation, since it is easier to balance.
2 HBr(aq) + ZnS(s) ⇄ H₂S(aq) + ZnBr₂(aq)
In the full ionic equation we include all ions and molecular species.
2 H⁺(aq) + 2 Br⁻(aq) + ZnS(s) ⇄ 2 H⁺(aq) + S²⁻(aq) + Zn²⁺(aq) + 2 Br⁻(aq)
In the net ionic equation we include only the ions that participate in the reaction and the molecular species.
ZnS(s) ⇄ S²⁻(aq) + Zn²⁺(aq)