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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s) + 3Cl2(g) -> 2AlCl3(s)

You are given 19.0g of aluminum and 24.0g of chlorine gas.

a)If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0g of aluminum?
= 0.704mol AlCl3

how do you do "b"

b)If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0g of chlorine gas, Cl2?
answer=_____mol AlCl3

  • CHEM -

    I don't understand how you can do a without knowing how to do b.
    mols Cl2 gas = 24.0/molar mass Cl2 = ??

    mols AlCl3 = mols Cl2 x (2 mols AlCl3/3 mols Cl2) =
    ??mols Cl2 x (2/3) = xx mols AlCl3.
    I hope this helps

  • CHEM -

    I kept trying it, via comparing it to a, & finally did come up with 0.255mol AlCl3...I had just gotten the question incorrect,somehow, I had to ask. Thank you for your patience with me. I will be looking into getting a tutor shortly.

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