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

Answer 1

Approx. #425*g#

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

Bravo for providing an equation that is stoichiometrically balanced.

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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Answer 2

Peyton Adams

To produce 4 moles of aluminum chloride, you would need 6 moles of chlorine gas. Since the balanced chemical equation shows a 2:3 mole ratio of aluminum to chlorine, you multiply the moles of aluminum chloride by the ratio to find the moles of chlorine needed. Then, you use the molar mass of chlorine (Cl2) to convert moles to grams.

6 moles Cl2 * (2 mol Cl2 / 3 mol AlCl3) * (70.906 g Cl2 / 1 mol Cl2) = 354.53 g Cl2

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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.

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

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