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Determine the mass of hydrogen chloride needed to react completely with 12.8 g of aluminum to produce aluminum chloride and hydrogen gas.

Respuesta :

first write out the eqn, and balance it.

next, take 12.8g/mr of aluminium
this will give u the mols of aluminium

next, take the ratio of HCL and Al, compare and find the mols of HCL.

take the mols of HCL/mr of HCL to give u mass

doneee
CintiaSalazar

Taking into account the reaction stoichiometry, 51.84 grams of HCl is required to react completely with 12.8 g of aluminum to produce aluminum chloride and hydrogen gas.

Reaction stoichiometry

In first place, the balanced reaction is:

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 2 moles
  • HCl: 6 moles
  • AlCl₃: 2 moles
  • H₂: 3 moles

The molar mass of the compounds is:

  • Al: 27 g/mole
  • HCl: 36.45 g/mole
  • AlCl₃: 133.35 g/mole
  • H₂: 2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al: 2 moles ×27 g/mole= 54 grams
  • HCl: 6 moles ×36.45 g/mole= 218.7 grams
  • AlCl₃: 2 moles ×133.35 g/mole= 266.7 grams
  • H₂: 3 moles ×2 g/mole= 6 grams

Mass of HCl required

The following rule of three can be applied: If by reaction stoichiometry 54 grams of Al react with 218.7 grams of HCl, 12.8 grams of Al react with how much mass of HCl?

[tex]mass of HCl=\frac{12.8 grams of Alx 218.7 grams of HCl}{54 grams of Al}[/tex]

mass of HCl= 51.84 grams

Finally, 51.84 grams of HCl is required to react completely with 12.8 g of aluminum to produce aluminum chloride and hydrogen gas.

Learn more about the reaction stoichiometry:

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