2 C6H6 + 15 O2 → 12 CO2 + 6 H2O ΔHrxn = 6542 kJ/rxn
If 8.88 g of C6H6 (FW = 78.114 g/mol) is burned and the heat produced from the reaction is used to heat 6354 g of water at 22.5o C, what is the final temperature of the water (swater = 4.184 J/g C)? (Write answer to 1 place past the decimal point)
two moles of benzene yield 6542kJ
so how many moles do you have? Ans 8.88/78.14
so the heat produced is 6542*(8.88/(2*78.14) kJ=371kj
heat produced=masswater*c*(Tf-22.5) solve for Tf
I get about a 14C rise in temp. Watch units, kj,kg,g
To find the final temperature of the water, we can use the equation:
q = m × C × ΔT
Where:
q = heat transferred (energy)
m = mass of the substance (in this case, water)
C = specific heat capacity of the substance (in this case, water)
ΔT = change in temperature
First, let's calculate the heat transferred from the combustion reaction using the given equation:
2 C6H6 + 15 O2 → 12 CO2 + 6 H2O ΔHrxn = 6542 kJ/rxn
However, we need to convert the heat produced (ΔHrxn) from kJ to J since the specific heat capacity of water (C) is given in J/g °C. Therefore:
ΔHrxn = 6542 kJ/rxn × 1000 J/kJ = 6542000 J/rxn
Now, we need to determine the number of moles of C6H6 used in the reaction:
molar mass of C6H6 = 78.114 g/mol
Given mass of C6H6 = 8.88 g
Number of moles of C6H6 = mass / molar mass
= 8.88 g / 78.114 g/mol
≈ 0.1137 mol
According to the balanced equation, the combustion of 2 moles of C6H6 produces 6 moles of H2O. Therefore, we can determine the number of moles of H2O produced in the combustion reaction:
Number of moles of H2O = (6/2) × 0.1137 mol
= 0.3411 mol
Next, let's calculate the heat transferred from the combustion reaction to the water. We can use the equation:
q = m × C × ΔT
Here, we know the mass of water (m = 6354 g) and the specific heat capacity of water (C = 4.184 J/g °C). We need to calculate the change in temperature (ΔT).
To do that, we can use the equation:
q = ΔHrxn
mwater × C × ΔT = ΔHrxn
ΔT = ΔHrxn / (mwater × C)
Substituting in the known values:
ΔT = 6542000 J/rxn / (6354 g × 4.184 J/g °C)
Now, let's calculate the final temperature (Tf) of the water using the equation:
Tf = Ti + ΔT
Where:
Ti = initial temperature of the water
ΔT = change in temperature
Given:
Ti = 22.5°C
Substituting the values into the equation:
Tf = 22.5°C + ΔT
This will give us our final answer.