Question

Сколько граммов бромида алюминия получится при реакции 135г. алюминия с бромом

Answer 1

270 грамм получиться после реакции(но это не точно,извени если не так)

Answer 2

Reaction Calculation: Aluminum Bromide Production

To calculate the amount of aluminum bromide produced from the reaction of 135g of aluminum with bromine, we can use the balanced chemical equation for the reaction.

The balanced chemical equation for the reaction between aluminum and bromine is: 2 Al + 3 Br2 -> 2 AlBr3

According to the stoichiometry of the reaction, 2 moles of aluminum react with 3 moles of bromine to produce 2 moles of aluminum bromide.

Using the molar mass of aluminum (Al) and bromine (Br), we can calculate the amount of aluminum bromide produced.

The molar mass of aluminum (Al) is approximately 26.98 g/mol, and the molar mass of bromine (Br) is approximately 79.904 g/mol.

Calculation:

1. Calculate the moles of aluminum: - Moles of aluminum = Mass of aluminum / Molar mass of aluminum - Moles of aluminum = 135g / 26.98 g/mol - Moles of aluminum ≈ 5.004 moles

2. Determine the moles of bromine required: - According to the balanced chemical equation, 3 moles of bromine are required for every 2 moles of aluminum. - Moles of bromine = (3/2) * Moles of aluminum - Moles of bromine ≈ (3/2) * 5.004 ≈ 7.506 moles

3. Calculate the mass of aluminum bromide produced: - Mass of aluminum bromide = Moles of aluminum bromide * Molar mass of aluminum bromide - Molar mass of aluminum bromide = (1 * molar mass of Al) + (3 * molar mass of Br) - Molar mass of aluminum bromide ≈ (1 * 26.98 g/mol) + (3 * 79.904 g/mol) ≈ 266.692 g/mol - Mass of aluminum bromide ≈ 7.506 moles * 266.692 g/mol ≈ 2003.5 grams

Therefore, approximately 2003.5 grams of aluminum bromide would be produced from the reaction of 135g of aluminum with bromine.