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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: 2Al(s)+3Cl2(g)→2AlCl3(s) You are given 14.0 gg of aluminum and 19.0 g of chlorine gas.   If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum? 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) You are given 14.0 gg of aluminum and 19.0 g of chlorine gas.   If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum? Express your answer to three significant figures and include the appropriate units.

Chemistry
Chemistry
10th Edition
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

You are given 14.0 gg of aluminum and 19.0 g of chlorine gas.

 

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 14.0 g of aluminum?
Express your answer to three significant figures and include the appropriate units.
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: 2A +B→A2B 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: 2.8 metA x 1 mol A2B 2 mełA = 1.4 mol A2B 3.2 mełB x 1 mol A2B 1 mełB = 3.2 mol A2B 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.
Transcribed Image Text: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: 2A +B→A2B 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: 2.8 metA x 1 mol A2B 2 mełA = 1.4 mol A2B 3.2 mełB x 1 mol A2B 1 mełB = 3.2 mol A2B 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.
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