Find the limiting reactant for the initial quantities of reactants
2k(s) + Cl2(g)--> 2KCl(s)
1.8 mol K; 1 mol Cl2
I'm lost
Ah, you're lost? Don't worry, I'll find you! As for the limiting reactant in this chemical equation, let's calculate it together.
First, we need to determine the number of moles of the reactants we have. Given that we have 1.8 mol of K and 1 mol of Cl2, we have an excess of K but a limited supply of Cl2.
Now, let's do some math to confirm this. The stoichiometric ratio in the balanced equation tells us that 2 moles of K react with 1 mole of Cl2 to produce 2 moles of KCl.
To find out how many moles of KCl we can produce, we use the limiting reactant. Since we have 1 mol of Cl2, we can only produce 2 moles of KCl. However, since we have 1.8 mol of K, we can produce 1.8 x 2 = 3.6 mol of KCl.
Therefore, the limiting reactant is Cl2 because it limits the amount of product formed. I hope that clarifies things for you!
To find the limiting reactant, you need to compare the mole ratios of the reactants to determine which one runs out first.
The balanced equation is:
2 K(s) + Cl2(g) → 2 KCl(s)
The mole ratio between K and Cl2 is 2:1.
Given:
Moles of K = 1.8 mol
Moles of Cl2 = 1 mol
To compare the moles of K and Cl2, we'll use the mole ratio of 2:1.
Moles of Cl2 needed = 2 * (moles of K) = 2 * (1.8 mol) = 3.6 mol
Since we only have 1 mol of Cl2, which is less than 3.6 mol needed, the Cl2 is the limiting reactant.
Therefore, Cl2 is the limiting reactant, and K is the excess reactant.
To find the limiting reactant, you need to determine which reactant is completely consumed (limiting) and which reactant is left over (excess) after the reaction takes place.
First, write a balanced chemical equation for the reaction:
2K(s) + Cl2(g) → 2KCl(s)
The coefficients in the balanced equation represent the mole ratio between reactants and products. In this case, the ratio of K to Cl2 is 2:1.
Now, compare the mole ratio of reactants in the balanced equation to the actual amounts given.
Given:
1.8 mol K
1 mol Cl2
Using the mole ratio, we can determine how many moles of Cl2 are required to react with 1.8 mol K:
1.8 mol K * (1 mol Cl2 / 2 mol K) = 0.9 mol Cl2
Since the amount of Cl2 (1 mol) is greater than the required amount (0.9 mol), Cl2 is not the limiting reactant. It is in excess.
Therefore, the limiting reactant is K, with a quantity of 1.8 mol.
Very simple. You just run the usual stoichiometry twice.
First you have 1.8 moles K.
moles KCl produced = 1.8 moles K x (2 moles KCl/2 moles K) = 1.8 moles KCl
Then you have 1 mole Cl2.
moles KCl produced = 1 mole Cl2 x (2 moles KCl/1 mole Cl2) = 2 moles KCl.
Obviously both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller value AND the reagent producing the smaller value is the limiting reagent.