0.357 mol Submit My Answers Give Up Correct By comparing your answers for Parts
ID: 550936 • Letter: 0
Question
0.357 mol Submit My Answers Give Up Correct By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces the lesser amount of product Part C Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A1 (s) + 3C12 (g)2AIC, (s) What is the maximum mass of aluminum chloride that can be formed when reacting 33.0 g of aluminum with 38.0 g of chlorine? Express your answer to three significant figures and include the appropriate units Hints Value Units Submit My Answers Give Up Continue 30Explanation / Answer
Molar mass of Al = 26.98 g/mol
mass of Al = 33.0 g
we have below equation to be used:
number of mol of Al,
n = mass of Al/molar mass of Al
=(33.0 g)/(26.98 g/mol)
= 1.223 mol
Molar mass of Cl2 = 70.9 g/mol
mass of Cl2 = 38.0 g
we have below equation to be used:
number of mol of Cl2,
n = mass of Cl2/molar mass of Cl2
=(38.0 g)/(70.9 g/mol)
= 0.536 mol
we have the Balanced chemical equation as:
2 Al + 3 Cl2 ---> 2 AlCl3
2 mol of Al reacts with 3 mol of Cl2
for 1.2231 mol of Al, 1.8347 mol of Cl2 is required
But we have 0.536 mol of Cl2
so, Cl2 is limiting reagent
we will use Cl2 in further calculation
Molar mass of AlCl3 = 1*MM(Al) + 3*MM(Cl)
= 1*26.98 + 3*35.45
= 133.33 g/mol
From balanced chemical reaction, we see that
when 3 mol of Cl2 reacts, 2 mol of AlCl3 is formed
mol of AlCl3 formed = (2/3)* moles of Cl2
= (2/3)*0.536
= 0.3573 mol
we have below equation to be used:
mass of AlCl3 = number of mol * molar mass
= 0.3573*1.333*10^2
= 47.6 g
Answer: 47.6 g
Feel free to comment below if you have any doubts or if this answer do not work
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