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