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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s)2Al(s)+3Cl2(g)→2AlCl3(s) You are given 21.0 gg of aluminum and 26.0 gg of chlorine gas. Part A: If you had excess chlorine, how many moles of of aluminum chloride could be produced from 21.0 gg of aluminum? Express your answer to three significant figures and include the appropriate units.   Part B: If you had excess aluminum, how many moles of aluminum chloride could be produced from 26.0 gg of chlorine gas, Cl2Cl2? Express your answer to three significant figures and include the appropriate units.

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s)2Al(s)+3Cl2(g)→2AlCl3(s) You are given 21.0 gg of aluminum and 26.0 gg of chlorine gas. Part A: If you had excess chlorine, how many moles of of aluminum chloride could be produced from 21.0 gg of aluminum? Express your answer to three significant figures and include the appropriate units.   Part B: If you had excess aluminum, how many moles of aluminum chloride could be produced from 26.0 gg of chlorine gas, Cl2Cl2? Express your answer to three significant figures and include the appropriate units.

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

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

You are given 21.0 gg of aluminum and 26.0 gg of chlorine gas.

Part A: If you had excess chlorine, how many moles of of aluminum chloride could be produced from 21.0 gg of aluminum?

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

 
Part B: If you had excess aluminum, how many moles of aluminum chloride could be produced from 26.0 gg of chlorine gas, Cl2Cl2?
Express your answer to three significant figures and include the appropriate units.
I Review I Constants I Periodic Table In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over: Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A +B→A2B 2Al(s) + 3Cl2 (g)→2AIC\3 (s) But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately: You are given 21.0 g of aluminum and 26.0 g of chlorine gas. Part A 1 mol A2B 2.8 met A × 1.4 mol A2B 2 mełA If you had excess chlorine, how many moles of of aluminum chloride could be produced from 21.0 g of aluminum? 3.2 metB x 1 mol A2B 3.2 mol A2B Express your answer to three significant figures and include the appropriate units. || 1 mełB • View Available Hint(s) Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of A2B can be formed from the given amounts. 画」? Value Units Submit Part B If you had excess aluminum, how many moles of aluminum chloride could be produced from 26.0 g of chlorine gas, Cl2 ?
Transcribed Image Text:I Review I Constants I Periodic Table In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2B without anything left over: Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A +B→A2B 2Al(s) + 3Cl2 (g)→2AIC\3 (s) But what if you're given 2.8 mol of A and 3.2 mol of B? The amount of product formed is limited by the reactant that runs out first, called the limiting reactant. To identify the limiting reactant, calculate the amount of product formed from each amount of reactant separately: You are given 21.0 g of aluminum and 26.0 g of chlorine gas. Part A 1 mol A2B 2.8 met A × 1.4 mol A2B 2 mełA If you had excess chlorine, how many moles of of aluminum chloride could be produced from 21.0 g of aluminum? 3.2 metB x 1 mol A2B 3.2 mol A2B Express your answer to three significant figures and include the appropriate units. || 1 mełB • View Available Hint(s) Notice that less product is formed with the given amount of reactant A. Thus, A is the limiting reactant, and a maximum of 1.4 mol of A2B can be formed from the given amounts. 画」? Value Units Submit Part B If you had excess aluminum, how many moles of aluminum chloride could be produced from 26.0 g of chlorine gas, Cl2 ?
Expert Solution
Step 1

The reaction is:

2Al(s)+3Cl2(g)→2AlCl3(s)

We will calculate both theoretical yield for Al and Cl2.

 

 

 

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