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?
2C6H6(l)+15O2(g)-->12CO2(G)+6H2O(g) :delta:Hrxn=-6278kj
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.
actually density=mass divided by volume
To solve this problem, we need to use the concept of stoichiometry and energy calculations. Here's how you can find the volume of benzene required to produce 1.5 x 10^3 kJ of heat:
Step 1: Determine the moles of heat produced
The given value of heat (1.5 x 10^3 kJ) is the change in enthalpy (ΔH) for the reaction. We can use this information to find the moles of heat produced by dividing the given value by the ΔH value per mole of reaction:
moles of heat = (1.5 x 10^3 kJ) / (-6278 kJ/mol)
Step 2: Convert moles of heat to moles of benzene
From the balanced equation, we can see that the stoichiometric ratio of benzene (C6H6) to heat (ΔH) is 2:12. This means that for every 2 moles of benzene, 12 moles of heat are produced. Hence, we have:
moles of benzene = (moles of heat) x (2 mol benzene / 12 mol heat)
Step 3: Convert moles of benzene to grams
To convert moles of benzene to grams, we need to use the molar mass of benzene (C6H6) which is given as 78.11 g/mol. Multiply the moles of benzene by the molar mass:
grams of benzene = (moles of benzene) x (molar mass of benzene)
Step 4: Convert grams of benzene to volume
Since the density of benzene (d) is given as 0.88 g/mL, we can use the following relationship to convert grams to volume (V):
volume of benzene = (grams of benzene) / (density of benzene)
Now, let's substitute the given values into the equations and calculate:
moles of heat = (1.5 x 10^3 kJ) / (-6278 kJ/mol) ≈ 0.2393 mol
moles of benzene = (0.2393 mol) x (2 mol benzene / 12 mol heat) ≈ 0.0399 mol
grams of benzene = (0.0399 mol) x (78.11 g/mol) ≈ 3.117 g
volume of benzene = (3.117 g) / (0.88 g/mL) ≈ 3.54 mL
Therefore, the volume of benzene required to produce 1.5 x 10^3 kJ of heat is approximately 3.54 mL.