The standard enthalpy of formation of Mn2O3 is −962.3 kJ/mol. How much heat energy is liberated when 6.5 grams of manganese are oxidized by oxygen gas to Mn2O3 at standard state conditions? Answer in units of kJ

I'm plugging that into my calculator and when I plug it into the answer box it keeps telling me that the answer is incorrect.

2Mn + (3/2)O2 ==> Mn2O3 + 962.3 kJ

So you have 962.3 kJ heat liberated by oxidizing 2*54.94 g Mn. You want to know heat liberated by oxidizing 6.5 g. That's
962.3 kJ x (6.5/2*54.94) = ?

To find the heat energy liberated when 6.5 grams of manganese (Mn) are oxidized to Mn2O3, we need to use the concept of molar mass and stoichiometry.

1. First, find the number of moles of manganese (Mn) in 6.5 grams using the molar mass of manganese.
The molar mass of Mn is 54.938 g/mol.
Number of moles = Mass / Molar mass
Number of moles of Mn = 6.5 g / 54.938 g/mol

2. Calculate the number of moles of Mn2O3 produced using the stoichiometric ratio of the balanced equation.
The balanced equation for the reaction is:

2 Mn + 3/2 O2 → Mn2O3

From the equation, we can see that 2 moles of Mn are required to produce 1 mole of Mn2O3.
So, the number of moles of Mn2O3 = (Number of moles of Mn) / 2.

3. Calculate the heat energy liberated using the given standard enthalpy of formation of Mn2O3.
The heat energy liberated (q) can be calculated using the equation:

q = (Number of moles of Mn2O3) * ΔHf

where ΔHf is the standard enthalpy of formation.
Remember to convert the answer to kJ by dividing by 1000.

Substituting the values into the equation, the calculation becomes:

Number of moles of Mn = 6.5 g / 54.938 g/mol
Number of moles of Mn2O3 = (Number of moles of Mn) / 2
Heat energy liberated (q) = (Number of moles of Mn2O3) * ΔHf / 1000

Now you can substitute the values you calculated into the equation to find the heat energy liberated.

To find the heat energy liberated when 6.5 grams of manganese (Mn) are oxidized, we'll follow these steps:

1. Calculate the number of moles of manganese (Mn) in 6.5 grams.
- To do this, we'll use the molar mass of manganese (Mn), which is 54.938045 g/mol.
- Number of moles of Mn = mass (g) / molar mass (g/mol)
- Number of moles of Mn = 6.5 g / 54.938045 g/mol

2. Write the balanced chemical equation for the reaction.
- The balanced chemical equation for the oxidation of Mn to Mn2O3 is:
4 Mn + 3 O2 -> 2 Mn2O3

3. Use the stoichiometry from the balanced equation to relate the moles of Mn to the moles of Mn2O3.
- From the balanced equation, we can see that 4 moles of Mn react to form 2 moles of Mn2O3.
- So, the moles of Mn2O3 can be calculated as follows:
Moles of Mn2O3 = (moles of Mn) * (2 moles of Mn2O3 / 4 moles of Mn)

4. Calculate the heat energy liberated using the enthalpy of formation of Mn2O3.
- The enthalpy of formation of Mn2O3 is given as -962.3 kJ/mol.
- Heat energy liberated = (moles of Mn2O3) * (enthalpy of formation of Mn2O3)

Now, let's do the calculations:

1. Calculate the number of moles of manganese (Mn):
Number of moles of Mn = 6.5 g / 54.938045 g/mol

2. Calculate the moles of Mn2O3 formed:
Moles of Mn2O3 = (moles of Mn) * (2 moles of Mn2O3 / 4 moles of Mn)

3. Calculate the heat energy liberated:
Heat energy liberated = (moles of Mn2O3) * (enthalpy of formation of Mn2O3)

Finally, convert the result to kilojoules (kJ) by dividing by 1000, since the enthalpy of formation is in kJ/mol.

Complete the calculations to find the answer.