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Sagot :
The Answer to the question is 133.34 grams approximately.
As Al is in excess so, complete HCl will consume in reaction.
Molar Mass of HCl = Atomic mass of H + Atomic mass of Cl
Molar Mass of HCl = 1 + 35.45 = 36.45
Number of moles of HCl which reacting will excess Al are
moles = given mass / molar mass
moles = 108 / 36.45 = 2.96 moles of HCl
So 2.96 moles of HCl are taking part in reaction which is approximately 3 moles
The chemical reaction is as follows as:
3 HCl + (excess) Al → AlCl₃ + 3 H
It is clear from the reaction that 3 moles of HCl are making 1 mole of AlCl₃
So in our reaction also 3 mole of HCl is taking part which mean that it will form 1 mole of AlCl₃ .
1 mole AlCl₃ = mass formed AlCl₃ / molar mass of AlCl₃
mass formed AlCl₃ = molar mass of AlCl₃
mass formed AlCl₃ = 133.34 grams
So, approximately 133.34 grams of AlCl₃ is formed when we react 108 grams of HCl with excess AlCl₃.
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