How many grams of butane must be burned to provide the heat needed to melt a 85.0-g piece of ice (at its melting point) and bring the resulting water to a boil (with no vapourization)? Assume a constant pressure of 1 bar and use the following data:
Hfo(H2O(g))=-241.83 kJ/mol, Hfo(O2(g))=0 kJ/mol, Hfo(CO2(g))=-393.5 kJ/mol, Hfo(C4H10(g))=-124.7 kJ/mol
Tfus(H2O)=0 oC, Tvap(H2O)=100 oC
Hfus(H2O)=6.01 kJ/mol, Cp(H2O(l))=75.291 J/mol/oC
2C4H10 + 13O2 ==> 8CO2 + 10H2O
Use the Hf information to find delta H (DH) for the reaction .
DHrxn = (n*DHf products) - (n*DHf reactants). That is the number of joules (or kJ but keep it all straight) available from the reaction as written.
Now, how much heat do you need.
You need to melt the ice.
mass ice x heat fusion = ?
You need to raise T of H2O from zero C to 100 C.
mass ice melt (85g) x specific heat water x 100 = ?
The sum of those two = amount of heat needed.
g C4H10 needed = 116g x (kJ needed/kJ rxn)
Check my work. It's late here.
Okay, so I got:
DHrxn = 8(395.5)+10(-241.83)-[2(-127.7)+13(o)
=-5326.9 kJ
mass(moles) ice x heat fusion 4.718mol(85/18.016)*6.01KJ/mol? =28.355Kj
Then I have 4.718mol *75.291J/mol/oC x 100= 35522.2938J
so sum together= 28.355kJ + 35.522 =
63.877kJ
then 2(58) x (63.877/-53269kJ))=0.1391g
I'm not sure if I did this right I may have went wrong in the first step, so if you could steer me in the right direction it would be appreciated,
Thanks.
I found one SMALL math error and one large math error.
You used DHfCO2 = -395.5 but your post (and my text) shows -393.5 but that makes only a small difference in the problem. . The larger error is the last step where you typed in 53269 and you omitted the decimal; it should be 5326.9(actually with the correction of the CO2 I have 5316.9 for this). That changes the answer by a factor of 10 so it is 1.39 g C4H10 needed.
How did you get g C4H10 needed??? I'm also working on this question
g C4H10 needed = 2(58) x (63.877/5316.9) = ?
thanks!
To solve this problem, we need to calculate the amount of heat required to melt the ice and heat the resulting water to its boiling point. Then, we can use the heat of combustion of butane to determine the mass of butane needed.
First, let's calculate the heat required to melt the ice. The heat required for the phase change from solid to liquid is given by the equation:
Q = mass × Hfus(H2O)
Given that the mass of the ice is 85.0 g and the heat of fusion for water is 6.01 kJ/mol, we need to convert the mass of ice to moles of water. The molar mass of water is 18.015 g/mol:
moles of water = mass of ice / molar mass of water
= 85.0 g / 18.015 g/mol
Next, we can calculate the heat required to melt the ice:
Q = moles of water × Hfus(H2O)
Since Hfus(H2O) is given in kJ/mol, we need to convert moles of water to moles of water:
moles of water = moles of water × 1000 kJ/1 kJ
Now, let's calculate the heat required to heat the resulting water to its boiling point. The heat required is given by the equation:
Q = mass × Cp(H2O) × ΔT
Given that the mass of the water is equal to the moles of water calculated earlier, the specific heat capacity of water is 75.291 J/mol/oC, and the change in temperature is from 0 oC to 100 oC:
Q = moles of water × Cp(H2O) × ΔT
= moles of water × 75.291 J/mol/oC × (100 oC - 0 oC)
Now, let's calculate the total heat required by adding the heat for the phase change and the heat to raise the temperature:
Q_total = Q_melting + Q_heating
Finally, we can use the heat of combustion of butane to determine the mass of butane needed. The heat of combustion of butane is given by:
Q_combustion = moles of butane × Hfo(C4H10(g))
We need to find the moles of butane, which can be calculated as:
moles of butane = Q_total / Hfo(C4H10(g))
Given that the heat of combustion of butane is -124.7 kJ/mol:
moles of butane = Q_total / -124.7 kJ/mol
The mass of butane is then given by:
mass of butane = moles of butane × molar mass of butane
Finally, you can perform the calculations to find the mass of butane needed to provide the necessary heat.