If you had excess chlorine, how many moles of of aluminum chloride could be produced from 25.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. View Available Hint(s) nothing nothing Part B If you had excess aluminum, how many moles of aluminum chloride could be produced from 30.0 g of chlorine gas, Cl2? Express your answer to three significant figures and include the appropriate units.

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jufsanabriasa jufsanabriasa · Answer 1 · 2019-10-06T03:51:21+02:00

Answer:

A. From 25,0 g of aluminium you will produce 0,927 moles of aluminium chloride

B. From 30,0 g of chlorine gas you will produce: 0,282 moles of aluminium chloride

Explanation:

a. The reaction that you have in this problem is:

3 Cl₂ + 2 Al → 2 AlCl₃

25,0 g of aluminium are:

25,0 ₓ[tex]\frac{1mole}{26,98g}[/tex] = 0,927 moles

As Al:AlCl₃ ratio is 1, from 25,0 g of aluminium you will produce: 0,927 moles of aluminium chloride

b. 30,0 g of chlorine gas are:

30,0 ₓ[tex]\frac{1mole}{70,90g}[/tex] = 0,423 moles of Cl₂

As Cl₂:AlCl₃ ratio is ³/₂:

0,423 moles of Cl₂ ₓ [tex]\frac{2 molesAlCl_3}{3 molesCl_2}[/tex] = 0,282 moles of aluminium chloride

Thus, from 30,0 g of chlorine gas you will produce: 0,282 moles of aluminium chloride

I hope it helps!

niharikadhawanarora niharikadhawanarora · Answer 2 · 2021-10-22T17:26:25+02:00

Part A: [tex]3 Cl_{2} + 2 Al[/tex]→[tex]2AlCl_{3}[/tex] , 250 g of aluminum is: 25× 1 mole÷ 26.98 g = 0.927 moles.
Part B: 30g of chlorine is: 30× 1mole÷ 70.9 g= 0.423 moles. As ratio of Cl is [tex]\frac{3}{2}[/tex], 0.423 moles of [tex]Cl_{2}[/tex] × [tex]\frac{2 moles AlCl_{3} }{3 moles Cl_{2} }[/tex] = 0.282 moles.

As a result, you'll get 0.282 moles of aluminum chloride from 30,0 g of chlorine gas.

Thus, A. You can make 0.927 moles of aluminum chloride from 25 g of aluminum.

B. You'll get 0.282 moles of aluminum chloride from 30 g of chlorine gas.

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