Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.
To determine the limiting reactant, we need to compare the amount of product that can be formed by each reactant.
First, we need to convert the given masses of chlorine gas (Cl2) and potassium (K) to moles.
To find the number of moles of chlorine gas:
Given mass of Cl2 = 7.00 g
Molar mass of Cl2 = 2 * atomic mass of chlorine = 2 * 35.45 g/mol = 70.9 g/mol
Number of moles of Cl2 = Given mass / molar mass = 7.00 g / 70.9 g/mol ≈ 0.099 mol
To find the number of moles of potassium:
Given mass of K = 5.00 g
Molar mass of K = 39.10 g/mol
Number of moles of K = Given mass / molar mass = 5.00 g / 39.10 g/mol ≈ 0.128 mol
Now, let's find the stoichiometric ratio of the reactants based on the balanced chemical equation.
Balanced equation: Cl2 + 2K → 2KCl
From the equation, we can see that for every 1 mole of Cl2, we need 2 moles of K.
Using the ratio, we can determine the number of moles required of each reactant:
Moles of K needed = 2 * moles of Cl2 = 2 * 0.099 mol ≈ 0.198 mol
Now, let's compare the moles of reactants:
- Moles of Cl2 available = 0.099 mol
- Moles of K available = 0.128 mol
Since we have more moles of potassium (0.128 mol) than the moles required (0.198 mol), potassium is the excess reactant.
To find the mass of the excess reactant remaining, we can calculate the mass of potassium left after the reaction.
Mass of excess K = (Moles of K available - Moles of K needed) * Molar mass of K
Mass of excess K = (0.128 mol - 0.198 mol) * 39.10 g/mol ≈ -0.07 mol * 39.10 g/mol ≈ -2.74 g.
However, we cannot have a negative mass, so the excess reactant mass is zero. Hence, there is no excess potassium remaining.
Therefore, potassium is the limiting reactant, and there is no excess reactant remaining.
To determine the limiting reactant and the mass of the excess reactant, we need to compare the given amounts of chlorine gas and potassium and calculate the amount of potassium chloride that can be formed from each reactant.
Step 1: Write the balanced equation for the reaction:
2 K + Cl₂ -> 2 KCl
Step 2: Calculate the number of moles for each reactant:
Given mass of chlorine gas = 7.00 g
Molar mass of chlorine gas (Cl₂) = 71 g/mol
Number of moles of chlorine gas = given mass / molar mass = 7.00 g / 71 g/mol = 0.099 moles
Given mass of potassium = 5.00 g
Molar mass of potassium (K) = 39 g/mol
Number of moles of potassium = given mass / molar mass = 5.00 g / 39 g/mol = 0.128 moles
Step 3: Determine the limiting reactant:
To determine the limiting reactant, compare the mole ratios of the reactants from the balanced equation. The reactant with a smaller mole ratio will be the limiting reactant. From the balanced equation, the mole ratio of K to Cl₂ is 2:1. Therefore, we need twice as many moles of K as Cl₂ to react completely.
The mole ratio of potassium to chlorine in the given amounts is:
(0.128 moles K) : (0.099 moles Cl₂) ≈ 1.29 : 1
Since the mole ratio is less than 2:1, it means that there are not enough moles of K to react with all the chlorine. Therefore, potassium is the limiting reactant.
Step 4: Calculate the amount of potassium chloride formed:
From the balanced equation, we know that 2 moles of K produce 2 moles of KCl. Therefore, the mole ratio of KCl to K is 1:1.
Number of moles of KCl formed = Number of moles of K (since there is a 1:1 mole ratio)
= 0.128 moles
Step 5: Calculate the mass of the excess reactant remaining:
To find the mass of the excess reactant remaining, we need to subtract the amount of K reacted from the initial amount of K.
Mass of excess K remaining = Initial mass of K - Mass of reacted K
= 5.00 g - (moles of K reacted x molar mass of K)
= 5.00 g - (0.128 moles x 39 g/mol)
= 5.00 g - 4.99 g
= 0.01 g
Therefore, the mass of the excess reactant (potassium) remaining is 0.01 g.
2molK + 1molClgas create 2molKCl
nK/nClgas=2/1
5/39:7/31=1.3
so Cl gas excess
mClgas(excess) = 7-(5*71/78) = 2.45g