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) ?
To find out how many kJ of heat are released in the combustion of 16.0 g of C6H6 (liquid benzene), we need to use the balanced equation and the molar mass of C6H6.
First, we'll calculate the moles of C6H6 using the given mass of 16.0 g and its molar mass.
Molar mass of C6H6 = 12.01 g/mol (C) + 1.01 g/mol (H) x 6 = 78.12 g/mol (C6H6)
Moles of C6H6 = Mass of C6H6 / Molar mass of C6H6
= 16.0 g / 78.12 g/mol
= 0.205 mol
Now, to calculate the heat released, we can use the stoichiometry of the balanced equation and the value of ΔH°.
From the balanced equation, we can see that 2 moles of C6H6 produce ΔH° of -6535 kJ.
Therefore, for 0.205 moles of C6H6, the heat released can be calculated using the following proportion:
(0.205 mol C6H6 / 2 mol C6H6) = (x kJ / -6535 kJ)
Simplifying this equation, we can solve for 'x':
x = (0.205 mol C6H6 / 2 mol C6H6) * -6535 kJ
x = -672 kJ
Therefore, approximately 672 kJ of heat are released in the combustion of 16.0 g of C6H6.