The theoretical yield is 160 g H₂O.
<em>Moles of H₂</em> = 18 g H₂ × (1 mol H₂/2.016 g H₂) = 8.93 mol H₂
<em>Moles of H₂O</em> = 8.93 mol H₂O × (2 mol H₂O/2 mol H₂) = 8.93 mol H₂O
<em>Theoretical yield</em> of H₂O = 8.93 mol H₂O × (18.02 g H₂O/1 mol H₂) = 160 g H₂O
Answer:
mass Na2SO4 = 14.3816 g
Explanation:
sln Na2SO4:
∴ V = 450 mL
∴ <em>C </em>=<em> </em>0.2250 mol/L
∴ Mw ≡ 142.04 g/mol..... from literature
⇒ mol Na2SO4 = (0.2250 mol/L)(0.450 L) = 0.10125 mol
⇒ mass Na2SO4 = (0.10125 mol)(142.04 g/mol) = 14.3816 g
Answer:
Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):
Being Keq:
![K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5Bfructose%5D%5BPi%5D%7D%7B%5BFructose-1-P%5D%7D)
Initial conditions:
![[Fructose-1-P]=0.2M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D0.2M)
![[Fructose]=0M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0M)
![[Pi]=0M](https://tex.z-dn.net/?f=%5BPi%5D%3D0M)
Equilibrium conditions:
![[Fructose-1-P]=6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D6.52%2A10%5E%7B-5%7DM)
![[Fructose]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)
![[Pi]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BPi%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)
Free-energy for T=298K (standard):
