27) What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to the following reaction?
2 C6H6(l) + 15 O2(g) �¨ 12 CO2(g) + 6 H2O(g) ƒ¢H�‹rxn = -6278 kJ
The reaction you show is for two moles benzene. Since the molar mass is 78.11 g/mole, then 2 moles is 2 x 79.11 = 156.22 grams. Thus 156.22 g benzene will produce 6278 kJ of heat. You want to produce 1500 kJ. Use proportions to determine the mass of benzene needed, then use density = mass x volume to convert to volume. Post your work if you get stuck.
37.326 g
To determine the volume of benzene (C6H6) required to produce 1.5 x 10^3 kJ of heat, we need to use the given molar mass of benzene and the molar ratio in the balanced equation.
1) Calculate the number of moles of heat produced:
Given: ΔH_rxn = -6278 kJ
Target: Heat produced = 1.5 x 10^3 kJ
To find the number of moles of heat produced, we divide the target heat by the ΔH_rxn:
Number of moles = Heat produced / ΔH_rxn
Number of moles = (1.5 x 10^3 kJ) / (-6278 kJ)
2) Find the number of moles of benzene:
According to the balanced equation, the molar ratio between benzene (C6H6) and heat is 2:6278.
Therefore, the number of moles of benzene is:
Number of moles of benzene = Number of moles of heat x (2/6278)
3) Convert moles to grams:
Given: Density of benzene (d) = 0.88 g/mL
Given: Molar mass of benzene = 78.11 g/mol
We can use the molar mass to convert moles to grams using the equation:
Grams = Moles x Molar mass
4) Convert grams to volume:
Given: Density of benzene (d) = 0.88 g/mL
We can use the density to convert grams to volume using the equation:
Volume = Grams / Density
Follow these steps to calculate the volume of benzene required.
To determine the volume of benzene required to produce 1.5 x 10^3 kJ of heat according to the given reaction, we need to use the concept of stoichiometry and the given molar mass and density of benzene.
First, let's calculate the moles of heat released using the given heat of reaction (∆H_rxn) and the given amount of heat (1.5 x 10^3 kJ):
moles of heat = (amount of heat)/(∆H_rxn)
moles of heat = (1.5 x 10^3 kJ)/(-6278 kJ) [converting kJ to J]
moles of heat = -0.239 mol (since the reaction releases heat, the value is negative)
Next, we need to determine the moles of benzene used in the reaction. From the balanced chemical equation, we can see that 2 moles of C6H6 are required for every 12 moles of CO2 produced.
moles of C6H6 = (moles of heat) x (moles of C6H6 in the balanced equation)/(moles of CO2 in the balanced equation)
moles of C6H6 = -0.239 mol x (2/12) [since the coefficient of C6H6 is 2 and the coefficient of CO2 is 12]
moles of C6H6 = -0.0398 mol
Now, we can calculate the mass of benzene used in the reaction using the moles of benzene and the molar mass of benzene.
mass of C6H6 = (moles of C6H6) x (molar mass of C6H6)
mass of C6H6 = -0.0398 mol x (78.11 g/mol)
mass of C6H6 = -3.11 g (the negative sign indicates that benzene is consumed)
Finally, we can calculate the volume of benzene required using the mass of benzene and the density of benzene.
volume of benzene = (mass of C6H6)/(density of C6H6)
volume of benzene = -3.11 g/(0.88 g/mL)
volume of benzene = -3.53 mL
Therefore, approximately 3.53 mL of benzene is required to produce 1.5 x 10^3 kJ of heat in the given reaction.