Limiting Reactant Procedure In the following chemical reaction, 2 mol of A will

Question

Limiting Reactant Procedure
In the following chemical reaction, 2 mol of A will react with 1 mol  of B  produce 1 mol  A2B without anything left over:
2A+B?A 2
But what if you're given 2.8 mol m 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:

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)+3Cl 2 (g)?2AlCl3(s)
You are given 31.0g g of aluminum and 36.0g g of chlorine gas.

Part A

If you had excess chlorine, how many moles of of aluminum chloride could be produced from 31.0g g of aluminum?

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

Part B
If you had excess aluminum, how many moles of aluminum chloride could be produced from 36.0g g of chlorine gas, Cl 2   m Cl_2?
Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

2Al(s) + 3Cl2(g) --> 2AlCl3(s)
31g.........36g .............? g

If this were actually a "limiting reactant" problem and you're looking for the mass of product, then compute the mass of AlCl3 from each given mass and the actual amount will be the lower of the two.

Find mass of AlCl3 as if Al was the limiting reactant:
31g Al x (1mol Al / 27g Al) x (2 mol AlCl3 / 2 mol Al) x (133.5g AlCl3 / 1 mol AlCl3) = 153 g AlCl3

Find mass of AlCl3 as if Cl2 was the limiting reactant:
36g Cl2 x (1mol Cl2 / 71g Cl2) x (2 mol AlCl3 / 3 mol Cl2) x (133.5g AlCl3 / 1 mol AlCl3) = 45 1g AlCl3

So, from the masses you included in your problem, you will produce 45.1 grams of AlCl3. And we must also conclude that chlorine is the limiting reactant, not the reactant in excess, as you've mentioned.

The answer to your question is given above. 153 grams of AlCl3 can be produced from 31 grams of Al and excess Cl2. But the catch is that 36 g of Cl2 is not an excess.

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