When 10.00 g of iron is burned in O2(g)to form Fe3O4 (s) , enough heat is generated to raise the temperature of 784 g of water from 18.0 C to 38.0 C.
Calculate the heat of formation of Fe3O4 (s) under these conditions in kJ/mol.
I got 1518 kJ/mol using that equation, but I still got it wrong. Did I do something wrong?
Let me see your data.
molesIronoxide= ...
specific heat constant....
0.04319 moles F3O4
Specific Heat Constant: 4.18 J/mol*C
Oh, we've got a hot chemistry problem here! Let me put on my lab coat, and I'll give it a shot.
To solve this problem, we need to calculate the heat released when burning 10.00 g of iron. Then, we'll convert that heat into kJ/mol of Fe3O4. Here we go!
First, we need to determine the heat released during the reaction. We can use the equation:
heat released = mass of water × specific heat capacity × change in temperature
Mass of water = 784 g
Specific heat capacity of water = 4.184 J/g°C (I know I know, we're speaking in J/g°C, but we'll convert to kJ/mol later!)
Change in temperature = 38.0°C - 18.0°C
Substituting the values:
heat released = 784 g × 4.184 J/g°C × 20.0°C
Now let's convert the heat released from J to kJ:
heat released = (784 g × 4.184 J/g°C × 20.0°C) / 1000
Multiplying and dividing gives us the heat released in kJ.
Alright, now we need to convert the heat released into kJ/mol of Fe3O4. For this, we need to know the molar mass of Fe3O4, which is 231.53 g/mol.
Finally, we divide the heat released in kJ by the number of moles:
heat of formation = heat released (in kJ) / moles of Fe3O4
And voila! You've got the heat of formation of Fe3O4 under these conditions in kJ/mol.
To calculate the heat of formation of Fe3O4 (s), we need to use the information given and apply the principles of thermodynamics. The equation for the reaction is:
4Fe(s) + 3O2(g) -> 2Fe2O3(s)
We can use the molar mass of Fe (55.85 g/mol) and the known specific heat capacity of water (4.18 J/g·°C) to determine the amount of heat transferred to the water.
First, we need to calculate the energy required to raise the temperature of water from 18.0 °C to 38.0 °C. We can use the equation Q = m * c * ΔT, where Q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Q = (784 g) * (4.18 J/g·°C) * (38.0 °C - 18.0 °C)
Q = 627,008 J
Next, we need to convert the energy from Joules to kilojoules by dividing by 1000:
Q = 627,008 J / 1000
Q = 627.008 kJ
Now, we can use the balanced equation for the combustion of iron to calculate the heat of formation of Fe3O4 (s). Since the reaction involves 4 moles of iron (Fe) to form 2 moles of Fe2O3 (iron(III) oxide), we can write:
Q = (627.008 kJ) / (4 moles of Fe) * (1 mole of Fe3O4 / 4 moles of Fe)
Q = 157.002 kJ/mol
Therefore, the heat of formation of Fe3O4 (s) under these conditions is 157.002 kJ/mol.
Assume you have one mole of FeO.Fe2O3
Find the mol mass of that.
Then find how many moles you have in 10g
heatformation= masswater*specificheat*20C/molesFeO.Fe2O3