Respuesta :
Answer is: -197,1 kJ heat is released in reaction.
Chemical reaction: 2Al + 3Cl₂ → 2AlCl₃ ΔH = -1408,4 kJ/mol.
m(Al) = 7,5 g.
n(Al) = m(Al) ÷ M(Al)
n(Al) = 7,5 g ÷ 27 g/mol
n(Al) = 0,28 mol.
0,28 mol : 2 mol = x : -1408,4 kJ/mol.
ΔH = -197,1 kJ.
n - amount of substance.
Chemical reaction: 2Al + 3Cl₂ → 2AlCl₃ ΔH = -1408,4 kJ/mol.
m(Al) = 7,5 g.
n(Al) = m(Al) ÷ M(Al)
n(Al) = 7,5 g ÷ 27 g/mol
n(Al) = 0,28 mol.
0,28 mol : 2 mol = x : -1408,4 kJ/mol.
ΔH = -197,1 kJ.
n - amount of substance.
The given above gives us the idea that for every 2 mole of the metal, aluminum, the heat released is equal to 1408.4 kJ. First, we need to determine the number of moles of the metal. The molar mass of aluminum is equal to 26.9815 g/mol.
n = (7.5 g)/(26.815 g/mol)= 0.280 moles
Then, multiply the calculated value by 1408.4/2.
H = 0.280 mols (1408.4/2) = 196.96 kJ
Answer: 196.96 kJ.
n = (7.5 g)/(26.815 g/mol)= 0.280 moles
Then, multiply the calculated value by 1408.4/2.
H = 0.280 mols (1408.4/2) = 196.96 kJ
Answer: 196.96 kJ.