Consider the following reaction: 2K(s) + Cl2(g) -> 2KCl(s). If 8 mol potassium reacts with 10 mol chlorine, how much chlorine will be left over after the reaction?
I really do not know how to go about solving this, I am very new to this unit. Thanks!
It takes 2K for each Cl2
so, 8 mol K will react with 4 mol Cl2
That leaves 6 moles of Cl2 left over
How could I be so clueless? Thank you so much!
Note that this is a limiting reagent problem. All of the K reacts (that's the limiting reagent) so it takes as much Cl2 as it needs and you have some Cl2 remaining.
To solve this question, you need to use the concept of stoichiometry. Stoichiometry involves using the balanced equation to determine the ratios of reactants and products in a chemical reaction.
First, let's write down the balanced equation for the reaction:
2K(s) + Cl2(g) -> 2KCl(s)
According to the balanced equation, 1 mole of Cl2 reacts with 2 moles of K, resulting in the formation of 2 moles of KCl.
Now, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To find the limiting reactant, we need to compare the number of moles of K and Cl2 provided in the question. According to the question, there are 8 mol of K and 10 mol of Cl2. Since the ratio of K to Cl2 in the balanced equation is 2:1, for every 2 moles of K, 1 mole of Cl2 is required.
Therefore, to use up all of the 8 mol of K, we would need 8/2 = 4 mol of Cl2.
Since we have 10 mol of Cl2 available, which is more than the 4 mol required, Cl2 is in excess in this reaction.
To find out how much Cl2 is left over after the reaction, we need to determine the amount of Cl2 that reacts with the 8 mol of K.
Since 1 mole of Cl2 reacts with 2 moles of K, the amount of Cl2 required to react completely with 8 mol of K is (8/2) = 4 mol.
Therefore, the remaining amount of Cl2 after the reaction is 10 mol - 4 mol = 6 mol.
Hence, there will be 6 moles of chlorine left over after the reaction.