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 2A1 (s) + 3C12 (g)2A1Clg (s) You are given 29.0 g of aluminum and 34.0 g of chlorine gas. Part A If you had excess chlorine, how many moles of of aluminum chloride could be produced from 29.0 g of aluminum? Express your answer to three significant figures and include the appropriate units. Hints ValueUnits Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Part B If you had excess aluminum, how many moles of aluminum chloride could be produced from 34.0 g of chlorine gas, Cl2? Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A)

Molar mass of Al = 26.98 g/mol

mass of Al = 29 g
molar mass of Al = 26.98 g/mol
mol of Al = (mass)/(molar mass)
= 29/26.98
= 1.0749 mol


According to balanced equation
mol of AlCl3 formed = moles of Al
= 1.0749 mol

Answer: 1.08 mol

B)

Molar mass of Cl2 = 70.9 g/mol

mass of Cl2 = 34 g
molar mass of Cl2 = 70.9 g/mol
mol of Cl2 = (mass)/(molar mass)
= 34/70.9
= 0.4795 mol

According to balanced equation
mol of AlCl3 formed = (2/3)* moles of Cl2
= (2/3)*0.4795
= 0.3197 mol

Answer: 0.320 mol

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