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For the reaction
2 Al + 6 HCl → Al2Cl6 + 3 H2
how much hydrogen can be formed when 60 g
of aluminum reacts with excess hydrochloric
acid?
Answer in units of g.

Respuesta :

CintiaSalazar

Answer:

6.67 grams of hydrogen can be formed when 60 g  of aluminum reacts with excess hydrochloric  acid.

Explanation:

The balanced reaction is:

2 Al + 6 HCl → Al₂Cl₆ + 3 H₂

By stoichiometry of the reaction (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
  • Al₂Cl₆: 1 mole
  • H₂: 3 moles

Being the molar mass:

  • Al: 27 g/mole
  • HCl: 36.45 g/mole
  • Al₂Cl₆: 266.7 g/mole
  • H₂: 2 g/mole

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
  • Al₂Cl₆: 1 mole* 266.7 g/mole= 266.7 grams
  • H₂: 3* 2 g/mole= 6 grams

You can apply the following rule of three: if 54 grams of aluminum forms 6 grams of hydrogen, 60 grams of aluminum will form how much mass of hydrogen?

[tex]mass of hydrogen=\frac{60 grams of aluminum*6 grams of hydrogen}{54 grams of aluminum}[/tex]

mass of hydrogen= 6.67 grams

6.67 grams of hydrogen can be formed when 60 g  of aluminum reacts with excess hydrochloric  acid.

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