The standard enthalpy of formation of Mn2O3 is −962.3 kJ/mol. How much heat energy is liberated when 6.4 grams of man- ganese are oxidized by oxygen gas to Mn2O3 at standard state conditions?
Answer in units of kJ
Well, let me calculate that for you, but first let me put on my oxygen goggles and transform into a math clown!
Now, to find out the heat energy liberated, we first need to calculate the number of moles of manganese.
To do that, we will use the molar mass of manganese (Mn), which is approximately 54.94 g/mol.
So, the number of moles of manganese is calculated as:
moles of Mn = mass of Mn (g) / molar mass of Mn (g/mol)
moles of Mn = 6.4 g / 54.94 g/mol
Now, let's move on to the balancing act of the chemical equation for the reaction:
2 Mn + 3/2 O2 --> Mn2O3
According to the balanced equation, 2 moles of manganese (Mn) are required to form 1 mole of manganese oxide (Mn2O3).
So, we can determine the number of moles of Mn2O3 produced:
moles of Mn2O3 = moles of Mn / 2
Now, let's get back to our numbers to find out how much heat is liberated:
heat energy liberated = moles of Mn2O3 * standard enthalpy of formation of Mn2O3
heat energy liberated = moles of Mn / 2 * -962.3 kJ/mol
Plug in the numbers and let the math clown do the calculations!
To calculate the heat energy liberated when 6.4 grams of manganese are oxidized to Mn2O3, we need to use the given enthalpy of formation and convert the mass of manganese to moles.
1. Calculate the moles of manganese (Mn):
Molar mass of Mn = 54.94 g/mol
moles of Mn = mass / molar mass
moles of Mn = 6.4 g / 54.94 g/mol
moles of Mn ≈ 0.1163 mol
2. Using the balanced chemical equation:
2 Mn + 3/2 O2 → Mn2O3
We can see that the stoichiometric ratio between Mn and Mn2O3 is 2:1. This means that for every 2 moles of Mn, we get 1 mole of Mn2O3.
3. Calculate the moles of Mn2O3 formed:
moles of Mn2O3 = 1/2 * moles of Mn
moles of Mn2O3 ≈ 1/2 * 0.1163 mol
moles of Mn2O3 ≈ 0.0582 mol
4. Calculate the heat energy:
heat energy = moles of Mn2O3 * enthalpy of formation
heat energy = 0.0582 mol * -962.3 kJ/mol (negative because energy is liberated)
heat energy ≈ -56.0 kJ
Therefore, approximately 56.0 kJ of heat energy is liberated when 6.4 grams of manganese are oxidized to Mn2O3 at standard state conditions.
To find the heat energy liberated when 6.4 grams of manganese are oxidized, we need to follow these steps:
1. Determine the number of moles of manganese (Mn) present:
To do this, we need to find the molar mass of manganese (Mn) and then use it to convert the given mass of manganese into moles.
The molar mass of Mn is 54.94 g/mol (grams per mole). Therefore, the number of moles of Mn can be calculated as:
Number of moles = mass / molar mass
= 6.4 g / 54.94 g/mol
2. Write the balanced equation for the oxidation of manganese by oxygen gas to form Mn2O3:
4 Mn + 3 O2 → 2 Mn2O3
3. Determine the change in enthalpy (ΔH) for the reaction:
The given value of the standard enthalpy of formation (ΔHf) of Mn2O3 is −962.3 kJ/mol. This represents the heat liberated when one mole of Mn2O3 is formed at standard state conditions.
Since the balanced equation shows that 4 moles of Mn are needed to react and form 2 moles of Mn2O3, the change in enthalpy for the reaction can be calculated as:
ΔH = (ΔHf) × (2 moles of Mn2O3 / 1 mole of Mn2O3)
= −962.3 kJ/mol × (2/4)
4. Calculate the heat energy liberated:
The heat energy liberated can be found by multiplying the change in enthalpy by the number of moles of Mn, as the stoichiometry of the balanced equation indicates that one mole of Mn releases ΔH amount of energy.
Heat energy liberated = ΔH × number of moles of Mn
Substitute the calculated values and solve:
Heat energy liberated = ΔH × number of moles of Mn
= (−962.3 kJ/mol × (2/4)) × (6.4 g / 54.94 g/mol)
Finally, calculate the value to get the answer in kJ.
We first write the balanced chemical reaction:
4 Mn + 3 O2 -> 2 Mn2O3 ; ΔH = -962.3 kJ/mol Mn2O3
We then get the number of moles Mn by dividing the given mass by molar mass of Mn. The molar mass of Mn is 54.94 g/mol (get a periodic table and search for it).
6.4 / 54.94 = 0.1165 mol Mn
From the reaction, for every 4 moles of Mn, 2 moles of Mn2O3 are produced. Thus,
0.1165 mol Mn * (2 mol Mn2O3 / 4 mol Mn) = 0.05825 mol Mn2O3
Finally, we multiply this to ΔH:
0.05825 mol Mn2O3 * -962.3 kJ/mol Mn2O3 = -56.05 kJ
Hope this helps :3