Explanation:
i think the answer is....b
Answer:
-1605.1 kJ
Explanation:
The enthalpy of formation is the enthalpy of the reaction that forms the substance only by its constituents, so, substances formed by one element, such as O2, in its ambient temperature phase, have an enthalpy of formation 0.
The enthalpy is a measure of how much heat the system contains, and so, the variation of it measures the heat lost (ΔH <0) or gained (ΔH >0), and for a reaction:
ΔHrxn = ∑ni*ΔHi products - ∑ni*Hi reactants
Where ni represents the coefficient of the substance, so by the data given:
ΔHrxn = [2*(-241.82) + (-393.5)] - [-74.6]
ΔHrxn = -802.54 kJ/ mol of CH4
Thus, the heat released is the enthalpy multiplied by the number of moles of CH4:
Q = -802.54*2
Q = -1605.1 kJ
Answer:
![\large \boxed{\text{2.52 mol Al}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5Ctext%7B2.52%20mol%20Al%7D%7D)
Explanation:
2Al₂O₃ ⟶ 4Al +3O₂
n/mol: 1.26
The molar ratio is 4 mol Al:2 mol Al₂O₃.
![\text{Moles of Al} = \text{1.26 mol Al$_{2}$O}_{3} \times \dfrac{\text{4 mol Al}}{\text{2 mol Al$_{2}$O}_{3}}= \textbf{2.52 mol Al}\\\\\text{The reaction produces $\large \boxed{\textbf{2.52 mol Al}}$}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Al%7D%20%3D%20%5Ctext%7B1.26%20mol%20Al%24_%7B2%7D%24O%7D_%7B3%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B4%20mol%20Al%7D%7D%7B%5Ctext%7B2%20mol%20Al%24_%7B2%7D%24O%7D_%7B3%7D%7D%3D%20%5Ctextbf%7B2.52%20mol%20Al%7D%5C%5C%5C%5C%5Ctext%7BThe%20reaction%20produces%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B2.52%20mol%20Al%7D%7D%24%7D)
The correct equilibrium expression would be one that has products over reactants, and to the power of their stoichiometric coefficient. Also, it would not include any liquids or solids.
The equilibrium expression for your balanced reaction is: