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

Answer 1

#234.9g#

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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Answer 2

Cooper Anderson

To find the mass of aluminum required to produce 8.70 moles of aluminum chloride (AlCl3), use the stoichiometry of the reaction. Since 2 moles of aluminum (Al) produce 2 moles of aluminum chloride (AlCl3), the molar ratio is 1:1. Therefore, the mass of aluminum required is equal to the molar mass of aluminum chloride multiplied by the number of moles of aluminum chloride produced. The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol. Thus, the mass of aluminum required is 8.70 moles × 133.34 g/mol = 1160.298 g.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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