19.9 g of aluminum and 235 g of chlorine gas react until all of the aluminum metal has been converted to AlCl₃. The balanced equation for the reaction is the following. [tex]2 Al(s) + 3Cl_2(g) \rightarrow 2AlCl_3(s)[/tex]What is the quantity of chlorine gas left, in grams, after the reaction has occurred, assuming that the reaction goes to completion? (The formula mass of aluminum metal, Al, is 26.98 g/mol, and the formula mass of chlorine gas, Cl₂, is 70.90 g/mol.)

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anfabba15 anfabba15 · Answer 1 · 2019-11-27T04:29:59+01:00

Answer:

The quantity of chlorine gas left, in grams, after the reaction has occurred is 156.2 g

Explanation:

2Al (s) + 3Cl₂ (g) → 2AlCl₃ (s)

First step: We should know the moles we have, of each reactant.

Mass / Molar weight = Moles

Moles Cl₂ : 235g / 70.9 g/m = 3.31 moles

Moles Al : 19.9 g/ 26.98 g/m = 0.73 moles

Now the equation:

2 moles of Al ___ react ___ 3 moles Cl₂

0.73 moles of Al ___ react ___ 1.10 moles Cl₂

(0.73 .3) / 2 = 1.10

I have 3.31 moles of Cl₂ and I only need 1.10 moles to complet the total reaction of Al.

3.31 moles - 1.10 moles = 2.21

These are the moles that remain to react.

Moles . molar weight = mass

2.21 moles . 70.9 g/m = 156.2 g