Course Home @ CAssignment 6 Chapter 4.3-4.4 Limiting Reactant Procedure session

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Course Home @ CAssignment 6 Chapter 4.3-4.4 Limiting Reactant Procedure session Chemis ) 20111 Aluminum reacts with chiorine gas to form aluminum chloride via the following reaction 2Al (s) + 3C1, (g)2A10%, (s) You are given 12.0 g of aluminum and 17.0 g of chlonine gas. In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of AzB without anything left over ork Part A 2A + BA,B you had excess chlorine, how many moles of of aluminum chloride could be produced from 12.0g of aluminum? 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, cakculate the amount of product formed from each amount of reactant separately Express your answer to three significant figures and include the appropriate units. dy View Available Hlnt(s) aring |Laluel Units 1 mol AgB 3.2 per × THA. 3.2 formed with mol AzB ,B 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 AzB can be formed from the given amounts Submit Part B if you had excess alumioum, how many moles of aluminum chloride could be produced from 17.0 g of chlorine gas. Cl2? Express your answer to three significant figures and include the appropriate units View Available Hints Value Units Submit

Explanation / Answer

Part a) if excess of chlorine is present then product moles is only depends upon moles of limiting reactant that is Aluminium....

Moles of aluminum= 12/27 = .444moles

According to stoichiometry,2 moles of aluminum chloride are produced if 2 moles of aluminum consumed provided chlorine is in excess amount...

So formed mole of aluminum chloride = .444 moles

Part b) moles of chlorine = 17/35.5 = .4788

For 3 moles of chlorine consumed formed mole of Aluminium chloride is 2 moles if aluminum is excess reactant...

So for 1 mole of chlorine 0.67 mole of aluminum chloride are produced

So for .4788 mole of chlorine formed mole of product is 0.4788*0.67= 0.3208 moles

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