In the following chemical reaction, 2 mol of A will react with 1 mol of B to pro

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: 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 roerfx lmolAiB 2 vp.Correct If you had excess aluminum, how many moles of aluminum chloride could be produced from 21.0 g of chlorine gas, CI2 ? Express your answer to three significant figures and include the appropriate units. 0.197 mol 1.4 mol A2B 1 mol A2B 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. By comparing your answers for Parts A and B, you can determine which reactant is limiting. Keep in mind that the limiting reactant is the one that produces the lesser amount of product. Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction: What is the maximum mass of aluminum chloride that can be formed when reacting 16.0 g of aluminum with 21.0 g of chlorine? Express your answer to three significant figures and include the appropriate units. Value Units

Explanation / Answer

Solution :-

Balanced reaction equation

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

16.0 g Al and 21.0 g Cl2 reacted what mass of Al2Cl3 will form

Using the balanced reaction mole ratio of the reactants and product we can calculate the mass of AlCl3 formed

Lets first calculate the moles of Al and Cl2

Mole =mass / molar mass

Moles of Al = 16.0 g / 26.982 g per mol = 0.593 mol Al

Moles of Cl2 = 21.0 g / 70.9 g per mol = 0.2962 mol Cl2

Now mole ratio of the Al to Cl2 is 2 :3

But here moles of the Cl2 are less than moles of Al

Therefore Cl2 is the limiting reactant

Now lets calculate the moles of the AlCl3 formed from the 0.2962 mol Cl2

0.2962 mol Cl2 * 2 mol AlCl3 / 3 mol Cl2 = 0.1975 mol AlCl3

Now lets convert moles of AlCl3 to its mass

Mass = moles * molar mass

Mass of AlCl3 = 0.1975 mol * 133.341 g per mol

                         = 26.3 g AlCl3

Therefore the mass of AlCl3 that can be produced = 26.3 g AlCl3

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