4Al(s) + 3O2(g) --> 2Al2O3(s) This is the balanced.
From the equation:
4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3
3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3 (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3
4.5 moles of O2 reacted with 6moles of Al to produce 3moles of Al2O3
(3) is the answer. 6 mol of Al.
C.) remove yourself from the radiation every 15 mins to prevent accumulation
Answer:
-255.4 kJ
Explanation:
The free energy of a reversible reaction can be calculated by:
ΔG = (ΔG° + RTlnQ)*n
Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.
C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)
Q = pC₂H₆/[pC₂H₂ * (pH₂)²]
Q = 0.261/[8.58*(3.06)²]
Q = 3.2487x10⁻³
ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)
ΔG = -255.4 kJ
I'm not so sure but I would say Answer Choice B
