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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 g of aluminum and 26.0 g of chlorine gas. What is the maximum mass of aluminum chloride that can be formed when reacting 21.0 g of aluminum with 26.0 g of chlorine? Express your answer to

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 g of aluminum and 26.0 g of chlorine gas. What is the maximum mass of aluminum chloride that can be formed when reacting 21.0 g of aluminum with 26.0 g of chlorine? Express your answer to

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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 g of aluminum and 26.0 g of chlorine gas.

What is the maximum mass of aluminum chloride that can be formed when reacting 21.0 g of aluminum with 26.0 g of chlorine?

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 metA x 1.4 mol A2B Part B 2 mełA 1 mol A2B 3.2 metB x 3.2 mol A2B Part C 1 mełB 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. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A1(s)+ 3C12 (9)¬→2AIC!3 (s) What is the maximum mass of aluminum chloride that can be formed when reacting 21.0 g of aluminum with 26.0 g of chlorine? Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) ? Value Units Submit
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 metA x 1.4 mol A2B Part B 2 mełA 1 mol A2B 3.2 metB x 3.2 mol A2B Part C 1 mełB 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. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2A1(s)+ 3C12 (9)¬→2AIC!3 (s) What is the maximum mass of aluminum chloride that can be formed when reacting 21.0 g of aluminum with 26.0 g of chlorine? Express your answer to three significant figures and include the appropriate units. • View Available Hint(s) ? Value Units Submit
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