the value of ΔH° for the reaction of 2 C6H6 (l) + 15 O2 (g) = 12 CO2 + 6H20 is -6535 kJ. how many kJ of heat are released in the combustion of 16.0 g of C6H6 (l) ?
heat released = 6535 kJ x (16.0/2*molar mass C6H6)
To find the amount of heat released during the combustion of 16.0 g of C6H6, we need to use the molar mass of C6H6 and the stoichiometry of the reaction. Here are the step-by-step calculations:
1. Calculate the molar mass of C6H6:
The molar mass of C6H6 is the sum of the atomic masses of carbon (C) and hydrogen (H) in one mole of C6H6.
Molar mass of C6H6 = (6 × atomic mass of C) + (6 × atomic mass of H)
Molar mass of C6H6 = (6 × 12.01 g/mol) + (6 × 1.01 g/mol)
Molar mass of C6H6 = 78.11 g/mol
2. Convert grams of C6H6 to moles:
To convert grams of C6H6 to moles, divide the given mass by the molar mass.
Number of moles of C6H6 = Mass of C6H6 / Molar mass of C6H6
Number of moles of C6H6 = 16.0 g / 78.11 g/mol
Number of moles of C6H6 ≈ 0.205 mol
3. Use the stoichiometry of the equation to find the heat released:
From the balanced equation, we see that 2 moles of C6H6 produce 6535 kJ of heat. So, we can set up a proportion to find the heat released when 0.205 moles of C6H6 are combusted.
Heat released (kJ) / 0.205 mol = -6535 kJ / 2 mol
Heat released (kJ) = (0.205 mol / 2 mol) × (-6535 kJ)
Heat released (kJ) ≈ -670.21 kJ
Therefore, approximately -670 kJ of heat is released during the combustion of 16.0 g of C6H6. The negative sign indicates that the reaction is exothermic, meaning heat is released.
To determine the amount of heat released in the combustion of 16.0 g of C6H6, we'll need to use stoichiometry and the given value of ΔH° (enthalpy change) for the reaction.
Here's how you can calculate it step by step:
1. Determine the molar mass of C6H6 (benzene).
The molar mass of C6H6 can be calculated by summing the atomic masses of its constituent elements:
C6H6 = (6*C) + (6*H) = 12.01 g/mol + 1.01 g/mol = 78.11 g/mol
2. Convert the given mass of C6H6 (16.0 g) to moles.
Moles = mass in grams / molar mass
Moles = 16.0 g / 78.11 g/mol ≈ 0.205 mol (rounded to three decimal places)
3. Use the stoichiometry of the balanced equation to determine the moles of CO2 produced.
According to the balanced equation: 2 C6H6 (l) + 15 O2 (g) = 12 CO2 + 6 H2O
2 moles of C6H6 are required to produce 12 moles of CO2.
Moles of CO2 produced = Moles of C6H6 * (12 mol CO2 / 2 mol C6H6)
= 0.205 mol * (12 / 2)
= 1.23 mol
4. Calculate the amount of heat released using the molar enthalpy change (ΔH°) for the reaction.
Heat released = Moles of CO2 * ΔH°
= 1.23 mol * (-6535 kJ/mol)
≈ -8034.05 kJ (rounded to two decimal places)
Therefore, approximately -8034.05 kJ of heat are released in the combustion of 16.0 g of C6H6.