For a particular isomer of C 8 H 18 , the combustion reaction produces 5099.5 kJ of heat per mole of C 8 H 18 ( g ) consumed, under standard conditions. C 8 H 18 ( g ) + 25 2 O 2 ( g ) ⟶ 8 CO 2 ( g ) + 9 H 2 O ( g ) Δ H ∘ rxn = − 5099.5 kJ / mol What is the standard enthalpy of formation of this isomer of C 8 H 18 ( g ) ?

the standard enthalpy of formation of this isomer of C₈H₁₈ (g) =  -375 kj/mol

Explanation:

The given combustion reaction

C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O   (1)

Heat of reaction or enthalpy of combustion = -5099.5 kj/mol

from equation (1)

     ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation                    of reactants)

Or,     - 5099.5   = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)](2)

Given ΔH⁰f(CO₂) = - 393.5 kj/mol  & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0

Using equation (2)

ΔH⁰f(C₈H₁₈) = -621 kj/mol

law of conservation of matter

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2