The enthalpies of formation of the compounds in the combustion of methane,

Upper C upper H subscript 4 (g) plus 2 upper O subscript 2 (g) right arrow upper C upper o subscript 2 (g) plus 2 upper H subscript 2 upper O (g).,

are CH4 (g): Delta. Hf = –74.6 kJ/mol; CO2 (g): Delta. Hf = –393.5 kJ/mol; and H2 O(g): Delta. Hf = –241.82 kJ/mol.

How much heat is released by the combustion of 2 mol of methane?

Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..

-80.3 kJ

-802.5 kJ

-1,605.1 kJ

-6,420.3 kJ

The amount of heat released by the combustion of 2 moles of methane = 1605.1 kJ

Option 3 is correct.

ΔH°(combustion of 2 moles of methane)

= -1,605.1 kJ

Explanation:

The balanced equation for the combustion of methane is presented as

CH₄ + 2O₂ → CO₂ + 2H₂O

The heat of formation for the reactants and products are

CH₄ (g): ΔHf = –74.6 kJ/mol;

CO₂ (g): ΔHf = –393.5 kJ/mol;

and H₂O(g): ΔHf = –241.82 kJ/mol.

ΔHf for O₂ = 0 kJ/mol

ΔH°rxn for the combustion of methane is given as

ΔH°rxn = ΣnH°(products) - ΣnH°(reactants)

ΣnH°(products) = (1×-393.5) + (2×-241.82)

= -877.14 kJ/mol

ΣnH°(reactants) = (1×-74.6) + (2×0)

= -74.6 kJ/mol

ΔH°rxn = -877.14 - (-74.6) = -802.54 kJ/mol

For 2 moles of methane, the heat of combustion = 2 moles × -802.54 kJ/mol

= -1,605.08 kJ = -1,605.1 kJ

Hope this Helps!!!

indium, and thallium have three electrons i think