Question

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 mol A×1 mol A2B2 mol A=1.4 mol A2B

3.2 mol B×1 mol A2B1 mol B=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.

Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:

2Al(s)+3Cl2(g)→2AlCl3(s)

You are given 20.0 g of aluminum and 25.0 g of chlorine gas.

Part A: If you had excess chlorine, how many moles of of aluminum chloride could be produced from 20.0 g of aluminum?

Part B: If you had excess aluminum, how many moles of aluminum chloride could be produced from 25.0 g of chlorine gas, Cl2?

Answer 1

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2 Al + 342_ 2 AIC'z msolar_imass of A1 = 27gr/mol Part A & Cl2=7gr/mol from reaction it is clear that - AC=133.59m/ma - 2x27 gm Al forms 2x 133:5 gm Alch so 20gm Ai forms 2x133.5 x2o an AICI 2x27 mass of Alliz form = 98.89 gm Algz. Part B a - from reaction it is clear that 3x7ign Cl2 efooms 2x133-5 gm AlCl3 so 25gm al fooms 2x133.5x25 gm AlCL 3*71. nass of AlCl3 forms = 47.01 gm Allz