18.7 g of aluminum and 226 g of chlorine gas react until all of the aluminum...

18.7 g of aluminum and 226 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. 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) 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.)

Expert Solution

2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)

no of moles of Al = W/G.A.Wt

= 18.7/27 = 0.69 moles

no of moles of Cl2 = W/G.A.Wt

= 226/71 = 3.183moles

2moles of Al react with 3 moles of Cl2

0.69 moles of Al react with = 3*0.69/2 = 1.035 moles of Cl2

Cl2 is exess reactant

the moles of chlorine gas left after complete the reaction = 3.183-1.035 = 2.148 moles

the quantity of chlorine gas left, in grams, after the reaction has occurred, = no of moles * gram molar mass

= 2.148*71 = 152.5g

queen_honey_blossom answered 2 years ago

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