solution is mixed with 40.0 mL of a 0.128 M lead(II) acetate solution and the following precipitation reaction occurs:

Question

solution is mixed with 40.0 mL of a 0.128 M lead(II) acetate solution and the following precipitation reaction occurs:

K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s)
The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g .

Identify the limiting reactant.
Pb(C2H3O2)2
PbSO4
K2SO4
KC2H3O2

My class has just started learning about limiting reactants and I do not understand how to break down the equation to find it. Could someone please explain the steps to find it?

Thanks!

Answer 1

find the final moles of PbSO4(s)

compare with the initial moles of Pb(C2H3O2)2(aq)

was all of the acetate converted?

if yes , then the acetate is the limiting reactant

if no , then the K2SO4(aq) is the limiting reactant

Answer 2

There are at least two ways for limiting reagent (LR) problems. I do them the long way. However, it appears to me you haven't posted the entire problem. What you have posted could be a LR problem but usually there are numbers for BOTH reactants. Please advise.

K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s)

Answer 3

To determine the limiting reactant, we need to compare the number of moles of each reactant and calculate how much product can be formed with each reactant. The reactant that produces the lowest amount of product is the limiting reactant.

Let's break down the equation and calculate the number of moles of each reactant:

1) Pb(C2H3O2)2(aq) + K2SO4(aq) -> 2KC2H3O2(aq) + PbSO4(s)

The molar mass of Pb(C2H3O2)2 is: 207.2 g/mol
The molar mass of K2SO4 is: 174.26 g/mol

To find the moles of each reactant, we can use the formula:

moles = mass / molar mass

For Pb(C2H3O2)2:
mass = volume x concentration = 40.0 mL x 0.128 M = 5.12 g
moles = 5.12 g / 207.2 g/mol = 0.0247 mol

For K2SO4:
No mass is given, so we will assume it is the limiting reactant and calculate moles based on the given volume and concentration.
moles = volume x concentration = 40.0 mL x 0.128 M = 5.12 g / 174.26 g/mol = 0.0294 mol

Now, let's calculate the number of moles of product formed for each reactant:

For PbSO4:
From the balanced equation, we can see that the stoichiometric coefficient of PbSO4 is equal to 1.
So, the moles of PbSO4 formed will be equal to the moles of the limiting reactant.

For KC2H3O2:
From the balanced equation, we can see that the stoichiometric coefficient of KC2H3O2 is equal to 2.
So, the moles of KC2H3O2 formed will be 2 times the moles of the limiting reactant.

Now, let's calculate the moles of product formed with each reactant:

Moles of PbSO4 formed = 0.0247 mol (since it is equal to the moles of the limiting reactant)

Moles of KC2H3O2 formed = 2 x 0.0247 mol = 0.0494 mol (since it is 2 times the moles of the limiting reactant)

Since the moles of PbSO4 formed are less than the moles of KC2H3O2 formed, we can conclude that Pb(C2H3O2)2 is the limiting reactant.

Therefore, the answer is: Pb(C2H3O2)2.