The limiting reactant between the reaction of 6 moles of oxygen and 12 moles of mercury is mercury (Hg)
<h3>Balanced equation </h3>
4Hg + O₂ —> 2Hg₂O
From the balanced equation above,
4 moles of Hg reacted with 1 mole of O₂
<h3>How to determine the limiting reactant</h3>
From the balanced equation above,
4 moles of Hg reacted with 1 mole of O₂
Therefore,
12 moles of Hg will react with = 12 / 4 = 3 moles of O₂
From the above calculation, we can see that only 3 moles of O₂ out of 6 moles given is required to react completely with 12 moles of Hg.
Thus, Hg is the limiting reactant and O₂ is the excess reactant .
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Which one I'm so confused
Answer:
A. Iodine
Explanation:
First 2 points describe it's physical properties. Even if we don't remember them, we can easily figure out the answer to be a non-metal. As, non-metals usually have lower bp and mp than metals.
Both Iodine and hydrogen exists as a 2 atom molecule in gas phase.
But only elements of group 17 (halogens) have 7 electrons in their outer shell.
Again, salts are usually ionic compounds, formed by atoms with a greater difference in their electronegativities. Hence, the one that forms bonds with a metal to form a salt readily can be a non- metal.
All these points indicate that the element must be iodine.
Answer:
a) The value of the
is 0.52.
b)Concentration of all species when equilibrium reestablishes:
![[CO_2] = 0.4748 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.4748%20M)
![[H_2] = 0.0198 M](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%200.0198%20M)
![[CO] = 0.0752 M](https://tex.z-dn.net/?f=%20%5BCO%5D%20%3D%200.0752%20M)
![[H_2O] =0.0652 M](https://tex.z-dn.net/?f=%5BH_2O%5D%20%3D0.0652%20M)
Explanation:
a) 
Equilibrium concentration of species :
![[CO] = 0.050 M, [H_2] = 0.045 M, [CO_2] = 0.086 M, and [H_2O] = 0.040 M](https://tex.z-dn.net/?f=%5BCO%5D%20%3D%200.050%20M%2C%20%5BH_2%5D%20%3D%200.045%20M%2C%20%5BCO_2%5D%20%3D%200.086%20M%2C%20and%20%5BH_2O%5D%20%3D%200.040%20M)
The expression of equilibrium constant is given as :
![\K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}](https://tex.z-dn.net/?f=%5CK_c%3D%5Cfrac%7B%5BCO%5D%5BH_2O%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D)

b)

Initially:
0.50 M 0.045 M 0.050 M 0.040 M
At equilibrium ;
(0.50-x) M (0.045-x) M (0.050+x) M (0.040+x) M
![K_c=\frac{[CO][H_2O]}{[CO_2][H_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5BH_2O%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D)

Solving fro x;
x = 0.0252
Concentration of all species when equilibrium reestablishes:
![[CO_2] = (0.50-x) M=(0.50-0.0252)M = 0.4748 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%20%280.50-x%29%20M%3D%280.50-0.0252%29M%20%3D%200.4748%20M)
![[H_2] = (0.045-x) M= (0.045-0.0252) M=0.0198 M](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%20%280.045-x%29%20M%3D%20%280.045-0.0252%29%20M%3D0.0198%20M)
![[CO] = (0.050+x) M=(0.050+0.0252)M = 0.0752 M](https://tex.z-dn.net/?f=%20%5BCO%5D%20%3D%20%280.050%2Bx%29%20M%3D%280.050%2B0.0252%29M%20%3D%200.0752%20M)
![[H_2O] = (0.040+x) M=(0.040+0.0252) M=0.0652 M](https://tex.z-dn.net/?f=%5BH_2O%5D%20%3D%20%280.040%2Bx%29%20M%3D%280.040%2B0.0252%29%20M%3D0.0652%20M)