How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction #2Al + 3Cl_2 -> 2AlCl_3#?

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How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction #2Al + 3Cl_2 -> 2AlCl_3#?

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Answer 1 · 2016-03-29T20:45:09

#234.9g#

Explanation:

The balanced chemical equation represents the mole ratio in which the chemicals combine, and in this case illustrates that #2 mol Al# produces #2molAlCl_3#, hence these 2 chemicals are in a #1:1# ratio.

Thus, to produce #8.70mol# aluminium chloride, it will require #8.70mol# aluminium.

But this quantity of #Al# has a mass in grams of

#m=nxxM_r#

#=8.70molxx27g//mol#

#=234.9g#

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How many grams of aluminum are required to produce 8.70 moles of aluminum chloride in the reaction #2Al + 3Cl_2 -> 2AlCl_3#?

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