In a reaction chamber, 3.0mol of aluminum is mixed with 5.3mol Cl2 and reacts. The reaction is described by the following balanced chemical equation.
2Al + 3Cl2 ==> 2AlCl3
a)Identify the limiting reagent for the reaction.
b)Calculate the number of moles or products formed
c)Calculate the number of moles of excess reagent remaining after the reaction.
To determine the limiting reagent, we need to compare the molar ratios of the reactants to the balanced chemical equation. The reactant that has a smaller mole ratio compared to the balanced equation will be the limiting reagent.
a) The molar ratio of Al to Cl2 is 2:3, which means for every 2 moles of Al, we need 3 moles of Cl2.
Since we have 3.0 moles of Al, we can calculate the required moles of Cl2:
3.0 mol Al * (3 mol Cl2 / 2 mol Al) = 4.5 mol Cl2
However, we only have 5.3 mol of Cl2, which is greater than the required amount of 4.5 mol. Therefore, Cl2 is in excess and Al is the limiting reagent.
b) From the balanced equation, we see that 2 moles of Al will react with 3 moles of Cl2 to produce 2 moles of AlCl3.
Since Al is the limiting reagent, we can use its mole amount to calculate the moles of AlCl3 formed:
3.0 mol Al * (2 mol AlCl3 / 2 mol Al) = 3.0 mol AlCl3
Therefore, 3.0 moles of Al will produce 3.0 moles of AlCl3.
c) Since Al is the limiting reagent, it will be fully consumed in the reaction. Therefore, there will be no moles of Al remaining after the reaction.
To calculate the moles of excess Cl2 remaining, we can subtract the moles of Cl2 used in the reaction from the initial moles of Cl2:
5.3 mol Cl2 - 4.5 mol Cl2 = 0.8 mol Cl2
Therefore, 0.8 moles of Cl2 will remain as excess after the reaction.
To identify the limiting reagent for the reaction, we need to compare the moles of each reactant to the stoichiometric ratio given in the balanced chemical equation. The reactant that has fewer moles compared to the stoichiometric ratio is the limiting reagent.
a) Let's compare the moles of aluminum (Al) and chlorine gas (Cl2) to determine the limiting reagent.
Moles of Al = 3.0 mol
Moles of Cl2 = 5.3 mol
According to the balanced chemical equation, the stoichiometric ratio of Al to Cl2 is 2:3. This means that 2 moles of Al react with 3 moles of Cl2.
By using this ratio, we can calculate the moles of Cl2 required to completely react with the given moles of Al:
Moles of Cl2 needed = (3.0 mol Al) x (3 mol Cl2 / 2 mol Al) = 4.5 mol Cl2
Since we have 5.3 mol of Cl2, which is greater than the moles needed (4.5 mol), Cl2 is in excess and Al is the limiting reagent for the reaction.
b) To calculate the number of moles of products formed, we use the stoichiometric ratio provided by the balanced chemical equation.
From the balanced equation: 2Al + 3Cl2 → 2AlCl3, we see that for every 2 moles of Al reacting, 2 moles of AlCl3 are formed.
Now, since Al is the limiting reagent, we can calculate the number of moles of AlCl3 formed:
Moles of AlCl3 = (3.0 mol Al) x (2 mol AlCl3 / 2 mol Al) = 3.0 mol AlCl3
Therefore, 3.0 moles of AlCl3 are formed in the reaction.
c) To calculate the moles of excess reagent remaining after the reaction, we subtract the moles of the limiting reagent consumed from the initial moles of the excess reagent. In this case, the excess reagent is Cl2.
Moles of Cl2 remaining = Moles of Cl2 - Moles of Cl2 needed
Moles of Cl2 remaining = 5.3 mol - 4.5 mol = 0.8 mol
Therefore, after the reaction, 0.8 mol of Cl2 will be remaining.
Most limiting reagent problems are worked the same way. Print these instructions and memorize them.
1. Write and balanced the equation. You have that.
2. USUALLY you are given grams and you must convert to moles by moles = grams/molar mass. This already has the moles given and you can skip this step.
3a. Using the coefficients in the balanced equation, convert 3.0 moles Al to moles of AlCl3.
3b. Same procedure but convert 5.3 moles Cl2 to moles AlCl3.
3c. It is quite likely that the answer to 3a and 3b will not be the same which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. USUALLY, the problem asks for grams here and you would calculate grams AlCl3 from the smaller value = moles x molar mass. That isn't needed here since they don't ask for it but this is useful information for the future.
5. To do part d, you have identified the limiting reagent. The other reagent is not limiting; therefore, there will be some that remains unreacted. To know the final value, use the coefficients (as in step 3 above) to determine the moles that react with ALL of the limiting reagent, subtract from the initial amount, the different is the amount remaining unreacted. If you wish (although the problem doesn't ask for it), you can multiply moles remaining x molar mass to arrive at the grams of the other reactant after the reaction.