Limiting Reactant Procedure 2A+BA2B 2.8 mol A×1 mol A2B2 mol A=1.4 mol A2B 3.2 m

Question

Limiting Reactant Procedure

2A+BA2B

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

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

Part A

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?

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

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Part B

If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0 g of chlorine gas, Cl2?

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

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Limiting Reactant Procedure

In the following chemical reaction, 2 mol of A will react with 1 mol of B to produce 1 mol of A2Bwithout anything left over:

2A+BA2B

But what if you're given 2.8 mol of A and 3.2 molof 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 19.0 g of aluminum and 24.0 g of chlorine gas.

Part A

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 19.0 g of aluminum?

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

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Part B

If you had excess aluminum, how many moles of aluminum chloride could be produced from 24.0 g of chlorine gas, Cl2?

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

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Explanation / Answer

(A)
Mass of Al = 19.0 g.

Molar mass of Al = 27.0 g/mol

Moles of Al = mass / molar mass = 19.0 / 27.0 = 0.704 mol

From the balanced equation,

2 mol of Al forms 2 mol of AlCl3

Hence, 0.704 mol of Al forms 0.704 mol of AlCl3

(B)

Mass of Cl2 = 24.0 g.

Molar mass of Cl2 = 71.0 g.

Moles of Cl2 = mass / molar mass = 24.0 / 71.0 = 0.338 mol

From the balanced equation

3 mol of Cl2 forms 2 mol of AlCl3

Then,

0.338 mol of Cl2 forms 0.338 * 2 / 3 = 0.225 mol of AlCl3

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