#2Al + 3Cl_2 -> 2AlCl_3# How many grams of chlorine are required to produce 4 moles of aluminum chloride?

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#2Al + 3Cl_2 -> 2AlCl_3# How many grams of chlorine are required to produce 4 moles of aluminum chloride?

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Answer 1 · 2016-07-07T15:27:06

Approx. #425*g#

Explanation:

#2Al(s) + 3Cl_2(g) rarr 2AlCl_3(s)#

You have given a stoichiometrically balanced equation, so bravo.

The equation explicitly tells us that #54*g# of aluminum metal reacts with #6xx35.45*g# #Cl_2# gas to give #266.7*g# of #"aluminum trichloride"#. From where am I getting these numbers?

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#2Al + 3Cl_2 -> 2AlCl_3# How many grams of chlorine are required to produce 4 moles of aluminum chloride?

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