Butane (c4h10) undergoes combustion in excess oxygen to generate gaseous carbon dioxide and water. given δh°f[c4h10(g)] = –124.7 kj/mol, δh°f[co2(g)] = –393.5 kj/mol, δh°f[h2o(g)] = –241.8 kj/mol, how much energy is released (kj) when 8.30 g of butane is burned?

The value of Δ H butane (g) = -124.7 kJ/mol

The value of Δ H CO2 (g) = -393.5 kJ/mol

The value of Δ H H2O (g) = -241.8 kJ/mol

Mass of butane, m = 8.30 gm

Molar mass of butane is 58 gm/mol

Consider the reaction,

C₄H₁₀ + 6.5 O₂ = 4CO₂ + 5H₂O

Calculating the value of Δ H° rxn:

ΔH°rxn = ∑nH° f (products) - ∑nH° f (reactants)

Substituting the values we get,

Δ H° rxn = 4 (-393.5) + 5 (-241.8) - (-124.7)

= -1574 -1209 + 124.7

= -2783 - 124.7

= -2658.3 kJ/mol

Now, calculate the number of moles of butane in 8.30 gm.

Number of moles = mass/molar mass

= 8.30 / 58

= 0.143 moles

Thus, the total energy released in the reaction is,

Q = number of moles × ΔH° rxn

= 0.143 × (2658.3)

= 380.14 kJ

Hence, the total heat released in the reaction is 380.14 kJ.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol