Aluminum reacts with chlorine gas to form aluminum chloride via the following re

Question

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)2AlCl3(s) You are given 33.0 g of aluminum and 38.0 g of chlorine gas.

A) If you had excess chlorine, how many moles of of aluminum chloride could be produced from 33.0 gof aluminum?

Express your answer to three significant figures and include the appropriate units.

B) If you had excess aluminum, how many moles of aluminum chloride could be produced from 38.0 gof chlorine gas, Cl2?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

We have to balance the reaction bellow.

2Al(s)+3Cl2(g)2AlCl3(s)

Given 33.0 g of aluminum and 38.0 g of chlorine gas.

(33.0 g Al * 1 mol / 26.982 g)*(2 mol AlCl3/2 mol Al)*(133.341g / 1 mol AlCl3)= 163 g AlCl

Answer: mass of AlCl3 produced from 33.0 g Al is 163 g AlCl3

B) If you had excess aluminum, how many moles of aluminum chloride could be produced from 38.0 g of chlorine gas, Cl2

Solution :- Using the mole ratio of the Cl2 and AlCl3 mass of the AlCl3 can be calculated

Mole ratio of the Cl2 to AlCl3 is 3 : 2

(38.0 g Cl2 *1mol /70.9 g)*(2 mol AlCl3/3 mol Cl2)*(133.341 g / 1 mol AlCl3) = 47.6 g AlCl3

Therefore 38.0 g Cl2 can produce 47.6 g AlCl3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.