how much heat will be liberated if 37.8 g of Fe2O3 is allowed to react with 18.7 g of aluminum.
Fe2O3(s) + 2Al(s)--> Al2O3(s) + 2Fe(s) + 851.5kJ
This is a combination limiting reagent problem with heat thrown in for good measure. What is the limiting reagent? How much Fe2O3 reacts?
851.5 kJ x (g Fe2O3 reacting/molar mass Fe2O3) = kJ heat released.
Thank you
I checked to see which one is the L.R, which it is Fe2O3, then i converted grams into moles and then converted those moles in to heat released and got -201.5 kJ
This is the first time i have been on here, if any one wants to confirm that is the correct answer, I would greatly appreciate it.
Thanks again
To calculate the amount of heat liberated in this reaction, you need to determine the limiting reactant first. The limiting reactant is the one that gets completely consumed, thus determining the maximum amount of product that can be formed.
1. Convert the given masses of Fe2O3 and aluminum into moles. To do this, divide the given mass by the molar mass of each substance.
- Moles of Fe2O3 = 37.8 g / molar mass of Fe2O3
- Moles of aluminum = 18.7 g / molar mass of aluminum
2. Determine the stoichiometric ratio between Fe2O3 and aluminum. According to the balanced equation, it is 1:2. This means that for every 1 mole of Fe2O3, you need 2 moles of aluminum to react completely.
3. Calculate the number of moles available for each reactant based on their stoichiometry.
- Moles of Fe2O3 available = Moles of Fe2O3 * (1 mole of aluminum / 2 moles of Fe2O3)
- Moles of aluminum available = Moles of aluminum
4. Identify the limiting reactant. Compare the moles calculated in step 3 for both reactants. The reactant with fewer moles is the limiting reactant because it determines the maximum amount of product formed.
- The limiting reactant is the one with the smaller moles available.
5. Calculate the moles of product formed based on the stoichiometry of the balanced equation. Using the stoichiometric ratio, determine the moles of product formed from the limiting reactant.
- Moles of Al2O3 formed = Moles of limiting reactant * (1 mole of Al2O3 / 2 moles of aluminum)
6. Calculate the heat liberated in the reaction using the moles of product formed and the given heat of reaction.
- Heat liberated = Moles of Al2O3 formed * (851.5 kJ / 2 moles of Al2O3)
Remember to use proper units and perform the necessary unit conversions throughout the calculations.
Note: Make sure to check the stoichiometry and molar masses of the substances involved to ensure accurate results.