Answer:
Explanation:The Î""G°′ of the reaction is −7.180 kJ·mol−1. Calculate the equilibrium constant for the reaction at 25 °
The balanced equation for the above reaction is;
CH₄ + 2O₂ ---> CO₂ + 2H₂O
Stoichiometry of CH₄ to O₂ is 1:2
The number of methane moles present - 1.44 g/ 16 g/mol = 0.090 mol
Number of oxygen moles present - 9.5 g/ 32 g/mol = 0.30 mol
If methane is the limiting reagent,
0.090 moles of methane react with 0.090x 2 = 0.180 mol
only 0.180 mol of O₂ is required but 0.30 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.
Number of moles of water that can be produced - 0.180 mol
Therefore mass of water produced - 0.180 x 18 g/mol = 3.24 g
Therefore mass of 3.24 g of water can be produced
Answer:
pH = 9.475
Explanation:
Hello there!
In this case, according to the basic ionization of the hydroxylamine:
The resulting equilibrium expression would be:
![Kb=\frac{[HONH_3^+][OH^-]}{[HONH_2]} =1.1x10^{-8}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHONH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BHONH_2%5D%7D%20%3D1.1x10%5E%7B-8%7D)
Thus, we first need to compute the initial concentration of this base by considering its molar mass (33.03 g/mol):
![[HONH_2]_0=\frac{1.34g/(33.03g/mol)}{0.500L} =0.0811M](https://tex.z-dn.net/?f=%5BHONH_2%5D_0%3D%5Cfrac%7B1.34g%2F%2833.03g%2Fmol%29%7D%7B0.500L%7D%20%3D0.0811M)
Now, we introduce 
as the reaction extent which provides the concentration of the hydroxyl ions to subsequently compute the pOH:
However, since Kb<<<<1, it is possible to solve for by easily neglecting it on the bottom to obtain:
![x=[OH^-]=\sqrt{1.1x10^{-8}*0.0811}= 2.99x10^{-5}](https://tex.z-dn.net/?f=x%3D%5BOH%5E-%5D%3D%5Csqrt%7B1.1x10%5E%7B-8%7D%2A0.0811%7D%3D%202.99x10%5E%7B-5%7D)
Thus, the pOH is:
And the pH:
Regards!
Answer:
ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹
Explanation:
The equation for the reaction is given as;
C₃H₈(g) + 5O₂(g) → 4H₂O(g) + 3CO₂(g)
In order to determine the entropy change, we have to use the entropy valuues for the species in the reaction. This is given as;
S°[C₃H₈(g)] = 269.9 J K⁻¹ mol⁻¹
S°[O₂(g)] = 205.1 J K⁻¹ mol⁻¹
S°[H₂O(g)] = 188.8 J K⁻¹ mol⁻¹
S°[CO₂(g)] = 213.7 J K⁻¹ mol⁻¹
The unit of entropy is J K⁻¹ mol⁻¹
Entropy change for the reaction is given as;
ΔS°reaction = ΔS°product - ΔS°reactant
ΔS°reaction = [(4 * 188.8) + (3 * 213.7)] - [269.9 + (5 * 205.1)]
ΔS°reaction = 100.9 J K⁻¹ (mol C₃H₈)⁻¹
