Aluminum reacts with chlorine to form aluminum chloride. The equation for this
reaction is 2 Al(s) + 3 Cl2(g) 2 AlCl3(s). What is the percentage yield for a reaction in which 33.5 g of aluminum is reacted with excess chlorine to produce 164.5 g of aluminum chloride?
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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)â2AlCl3(s) You are given 26.0 g of aluminum and 31.0 g of chlorine gas.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 26.0 g of aluminum?
If you had excess aluminum, how many moles of aluminum chloride could be produced from 31.0 g of chlorine gas, Cl2?
Posted Dec 17 2019
AnsweredDec 17 2019
OC2735112
University of Toronto Scarborough
1)
Molar mass of Al = 26.98 g/mol
mass of Al = 26 g
mol of Al = (mass)/(molar mass)
= 26/26.98
= 0.9637 mol
According to balanced equation
mol of AlCl3 formed = moles of Al
= 0.9637 mol
Answer: 0.964 mol
2)
Molar mass of Cl2 = 70.9 g/mol
mass of Cl2 = 31 g
mol of Cl2 = (mass)/(molar mass)
= 31/70.9
= 0.4372 mol
According to balanced equation
mol of AlCl3 formed = (2/3)* moles of Cl2
= (2/3)*0.4372
= 0.2915 mol
Answer: 0.292 mol