An aqueous solution containing 9.23 g of lead(II) nitrate is added to an aqueous solution containing 7.20 g of potassium chloride.
How many grams of the excess reactant remain?
mols Pb(NO3)2 = 9.23/331.2 = 0.0279
mols KCl = 7.2/74.6 = .0965
..........Pb(NO3)2 + 2KCl ==> PbCl2(s) + 2KNO3
I..........0.0279........0.0965........0..................0
C........-0.0279......-2*0.0279 ....0.0279.......
E.............0............-0.0407........0.0279
If this is a lower level chemistry class, and I assume that is so, then 0.0407 mols KCl remain and that is the excess reagent. Convert to grams by grams = mols x molar mass = ?
To determine the grams of the excess reactant remaining, we first need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in a chemical reaction, limiting the amount of product formed.
Let's calculate the moles of lead(II) nitrate (Pb(NO3)2) and potassium chloride (KCl) separately.
Molar mass of Pb(NO3)2:
Pb: 207.2 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3)
Total: 207.2 + 14.01 + (16.00 x 3) = 331.2 g/mol
Molar mass of KCl:
K: 39.10 g/mol
Cl: 35.45 g/mol
Total: 39.10 + 35.45 = 74.55 g/mol
Now, let's calculate the moles of each reactant:
Moles of Pb(NO3)2 = mass / molar mass
Moles of Pb(NO3)2 = 9.23 g / 331.2 g/mol ≈ 0.0279 mol
Moles of KCl = mass / molar mass
Moles of KCl = 7.20 g / 74.55 g/mol ≈ 0.0966 mol
Based on the balanced chemical equation for the reaction between Pb(NO3)2 and KCl, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KCl to form 1 mole of PbCl2. Therefore, the ratio of moles between Pb(NO3)2 and KCl is 1:2.
Since the ratio is 1:2, it means that 0.0279 mol of Pb(NO3)2 will react with twice that amount of KCl, which is 0.0558 mol (2 x 0.0279 mol).
Now, let's calculate the amount of KCl remaining:
Moles of KCl remaining = Moles of KCl initially - Moles of KCl reacted
Moles of KCl remaining = 0.0966 mol - 0.0558 mol = 0.0408 mol
Finally, let's calculate the mass of the excess reactant:
Mass of excess reactant = Moles of excess reactant x Molar mass
Mass of excess reactant = 0.0408 mol x 74.55 g/mol ≈ 3.05 g
Therefore, approximately 3.05 grams of the excess reactant (potassium chloride) remain.
To determine the grams of the excess reactant remaining, we first need to identify the limiting reactant between lead(II) nitrate (Pb(NO3)2) and potassium chloride (KCl). The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
We can calculate the number of moles of each reactant using their respective molar masses. The molar mass of lead(II) nitrate (Pb(NO3)2) is 331.20 g/mol, and the molar mass of potassium chloride (KCl) is 74.55 g/mol.
1. Calculate the number of moles of lead(II) nitrate:
Moles of Pb(NO3)2 = mass / molar mass
Moles of Pb(NO3)2 = 9.23 g / 331.20 g/mol = 0.0279 mol
2. Calculate the number of moles of potassium chloride:
Moles of KCl = mass / molar mass
Moles of KCl = 7.20 g / 74.55 g/mol = 0.0966 mol
Now, we need to compare the moles of each reactant based on their stoichiometric ratio in the balanced equation. The balanced equation for the reaction between lead(II) nitrate and potassium chloride is:
Pb(NO3)2 + 2KCl → PbCl2 + 2KNO3
The mole ratio between Pb(NO3)2 and KCl is 1:2. Therefore, for every 1 mole of Pb(NO3)2, we need 2 moles of KCl.
3. Determine the stoichiometric ratio:
Moles of Pb(NO3)2 : Moles of KCl
0.0279 mol : 0.0966 mol
The ratio is approximately 1:3.47. This tells us that the stoichiometry does not allow for a 1:1 reaction between the two reactants. Therefore, the reactant in excess will be the one with a higher number of moles.
Since there is an excess of KCl, we can calculate the moles and grams of the excess reactant remaining:
4. Calculate the moles of KCl required for the reaction (based on Pb(NO3)2):
Moles of KCl required = 0.0279 mol Pb(NO3)2 × 2 mol KCl / 1 mol Pb(NO3)2 = 0.0559 mol KCl
5. Calculate the excess moles of KCl:
Excess moles of KCl = Moles of KCl initially - Moles of KCl required
Excess moles of KCl = 0.0966 mol - 0.0559 mol = 0.0407 mol
Finally, we can convert the excess moles of KCl to grams:
6. Calculate the excess grams of KCl:
Excess grams of KCl = Excess moles of KCl × molar mass of KCl
Excess grams of KCl = 0.0407 mol × 74.55 g/mol = 3.03 g
Therefore, 3.03 grams of the excess reactant (potassium chloride) will remain after the reaction.