Many cigarette lighters contain liquid butane, C4H10(l). Using standard enthalpies of formation, calculate the quantity of heat produced when 2.7 g of butane is completely combusted in air under standard conditions.
I assume you have a set of tables.
delta Hrxn = (delta Hproducts)-(delta H reactants).
Then delta Hrxn x (2.7/molar mass C4H10).
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To calculate the quantity of heat produced when butane is combusted, we need to use the standard enthalpies of formation (∆Hf) for butane and its combustion products (CO2 and H2O).
Here are the relevant standard enthalpies of formation (∆Hf) values:
∆Hf(C4H10(l)) = -126.77 kJ/mol
∆Hf(CO2(g)) = -393.51 kJ/mol
∆Hf(H2O(l)) = -285.83 kJ/mol
First, we need to find the molar mass of butane (C4H10):
Molar mass of C4H10 = (4 * Atomic mass of C) + (10 * Atomic mass of H)
= (4 * 12.01 g/mol) + (10 * 1.01 g/mol)
= 58.12 g/mol
Next, we calculate the number of moles of butane:
n(moles) = mass (g) / molar mass (g/mol)
= 2.7 g / 58.12 g/mol
≈ 0.0465 mol
Now, using the balanced chemical equation for the combustion of butane:
C4H10(l) + 13/2 O2(g) -> 4 CO2(g) + 5 H2O(l)
We can see that 1 mole of butane produces 4 moles of CO2. So, the moles of CO2 produced can be calculated as:
moles of CO2 = 4 * n(moles of butane)
= 4 * 0.0465 mol
= 0.186 mol
Similarly, we can calculate the moles of H2O produced:
moles of H2O = 5 * n(moles of butane)
= 5 * 0.0465 mol
= 0.233 mol
Now, we can calculate the heat produced by the combustion using the following equation:
∆H(combustion) = (moles of CO2 * ∆Hf(CO2)) + (moles of H2O * ∆Hf(H2O)) - (moles of butane * ∆Hf(C4H10))
Substituting the values:
∆H(combustion) = (0.186 mol * -393.51 kJ/mol) + (0.233 mol * -285.83 kJ/mol) - (0.0465 mol * -126.77 kJ/mol)
Calculating each term:
∆H(combustion) ≈ -73.12 kJ + -66.66 kJ + 5.90 kJ
≈ -133.88 kJ
Therefore, the quantity of heat produced when 2.7 g of butane is completely combusted is approximately 133.88 kJ.