How much heat (in kilojoules) is released on reaction of 4.10 g of Al?
Reaction: 2Al(s) + 3Cl2(g)--->2AlCl3(s)
Delta H= -1408.4 KJ
4.10 g Al = how many mols?
4.10/27 (look up the value) = ??
You get 1408.4 kJ/mol so how much will you get for ?? mole?
Didn't you post a question earlier asking how to prepare a solution of 471 g acetone + solvent (the question was for the solvent)? I posted an answer and now I can't find the question or the answer.
no i didn not. but thank you!
To determine the amount of heat released on the reaction of 4.10 g of Al, we need to use the stoichiometry of the reaction and the given molar enthalpy change (delta H).
First, we need to convert the mass of Al (4.10 g) into moles. We can do this by dividing the mass by the molar mass of aluminum (Al), which is 26.98 g/mol.
Moles of Al = Mass of Al / Molar mass of Al
Moles of Al = 4.10 g / 26.98 g/mol ≈ 0.152 moles
From the balanced equation, we can see that the stoichiometric ratio between Al and delta H is 2: -1408.4 KJ. This means that for every 2 moles of Al, -1408.4 KJ of heat is released.
Now we can determine the amount of heat released for 0.152 moles of Al:
Heat released = (-1408.4 KJ / 2) * 0.152 moles
Heat released = -107.0912 KJ
Therefore, on the reaction of 4.10 g of Al, approximately 107.0912 KJ of heat is released.