# 60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead () nitrate. What is the limiting reactant? How many grams of precipitate form?

The limiting reactant is potassium iodide (KI) and the mass of precipitate (PbI₂) formed is 4453.36 g

From the question,

We are to determine the limiting reactant in the reaction between the given potassium iodide and lead nitrate

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

This means

2 moles of KI will react with 1 mole of Pb(NO₃)₂ to produce 2 moles of KNO₃ and 1 mole of PbI₂

To determine the limiting reactant,

First, we will determine the number of moles of each reactant present

• For potassium iodide (KI)

Volume = 60.0 mL

Concentration = 0.322 M

From the formula

Number of moles = Concentration × Volume

∴ Number of moles of KI present = 0.322 × 60.0

Number of moles of KI present = 19.32 moles

Volume = 20.0 mL

Concentration = 0.530 M

∴ Number of moles of Pb(NO₃)₂ present = 0.530 × 20.0

Number of moles of Pb(NO₃)₂ present = 10.60 moles

Now, from the balanced chemical equation

2 moles of KI is required to react with 1 mole of Pb(NO₃)₂

Then,

19.32 moles of KI will react with moles of Pb(NO₃)₂

∴ 19.32 moles of KI will react with 9.66 moles of Pb(NO₃)₂

The number of moles of Pb(NO₃)₂ present is 10.60 moles.

Since this is more than the quantity that reacted (9.66 moles), then Pb(NO₃)₂ is the excess reactant and KI is the limiting reactant.

∴ KI is the limiting reactant

To determine how many grams of the precipitate formed,

The precipitate formed is lead iodide (PbI₂)

That is, we will determine the mass of lead iodide formed

From the balanced equation for the reaction,

2 moles of KI will react with 1 mole of Pb(NO₃)₂ to produce 2 moles of KNO₃ and 1 mole of PbI₂

Then,

19.32 moles of KI will react with 9.66 moles of Pb(NO₃)₂ to produce 19.32 moles of KNO₃ and 9.66 moles of PbI₂

Therefore, 9.66 moles of PbI₂ was formed during the reaction.

Now, for the mass of PbI₂ formed

From the formula

Mass = Number of moles × Molar mass

Number of moles of PbI₂ = 9.66 moles

Molar mass of PbI₂ = 461.01 g/mol

∴ Mass of PbI₂ formed = 9.66 × 461.01

Mass of PbI₂ formed = 4453.3566 g

Mass of PbI₂ formed ≅ 4453.36 g

Hence, the limiting reactant is potassium iodide (KI) and the mass of precipitate (PbI₂) formed is 4453.36 g

Limiting reagent: Potassium iodide

Mass of the precipitate (PbI₂) is 4.453 g

Explanation:

We are given;

• 60.0 mL of 0.322 M potassium iodide
• 20.0 mL of 0.530 M lead () nitrate

We are required to identify the limiting reactant and determine the mass of the precipitate formed.

Step 1: Write the balanced equation for the reaction
• The balanced equation for the reaction between potassium iodide and lead (II) nitrate is given by;

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂(s)

Step 2: Determine the number of moles of the reagents

Moles of KI

Moles = Molarity × volume

Moles of KI = 0.322 M × 0.060 L

= 0.01932 moles

Moles of KNO₃

Moles = 0.530 M × 0.020 L

= 0.0106 M

From the equation;

• 2 moles of KI reacts with 1 mole of Pb(NO)₂
• Therefore; 0.01932 moles of KI will require 0.00966 moles of Pb(NO₃)₂
• This means, KI is the limiting reagent while Pb(NO₃)₂ is the excess reagent.
Step 3: Determine the mass of the precipitate PbI₂

2 moles of KI reacts to produce 1 mole of PbI₂

Therefore;

Moles of PbI₂ = Moles of KI ÷ 2

= 0.01932 moles ÷ 2

= 0.00966 moles

But molar mass of PbI² is 461.01 g/mol

Therefore;

Mass of PbI₂ = 0.00966 moles × 461.01 g/mol

= 4.453 g

Therefore, the mass of the precipitate formed (Pbi₂)is 4.453 g

## Related Questions

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 4.0 at 228°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl5(g) is 0.26 M, what is the equilibrium concentration of PCl3?

Answer : The equilibrium concentration of is, 1.0 M

Explanation : Given,

Equilibrium concentration of = 0.26 M

Volume of solution = 1.00 L

Equilibrium constant = 4.0

The balanced equilibrium reaction will be,

The expression of equilibrium constant for the reaction will be:

From the reaction we conclude that the concentration of and are equal.

Let the concentration of be 'X'.

So, concentration of = X

Now put all the values in this expression, we get :

Thus,

The concentration of at equilibrium = X = 1.0 M

The concentration of at equilibrium = X = 1.0 M

HELP PLEASE! What is paper chromatography is and what it is used for? please explain as simple as possible.

Paper chromatography is used to separate pigments (color) as well as primary (Blue,Red,Yellow) and secondary (Purple,Orange,Green) inks for logical reasonings.

Hope this helps

What does Newton’s second law of motion state ?

The Newtons second law of motion states that"The rate of change of the momentum of a moving body is directly proportional to the force applied."

Derivation from,

F = △P/t

to,

F=ma

#SPJ4

E. solid copper sulfide and silver nitrate react to form copper (ii) nitrate and solid silver sulfide. write a balanced chemical equation that describes the reaction. identify the oxidation number of each element in the reaction. (you do not need to include the total contribution of charge.) is this reaction a redox reaction or a non-redox reaction? explain your answer.

The given reaction is not a redox reaction.

Explanation:

Redox reaction is defined as chemical reaction in oxidation and reduction reactions occurs simultaneously.

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, oxidation state of the atom decreases.

Solid copper sulfide when immersed in silver nitrate solution it reacts to form silver sulfide precipitate and copper(II) nitrate solution.

Oxidation states of elements on reactant side:

Oxidation state of copper  = +2

Oxidation state of sulfur = -2

Oxidation number of silver  +1

Oxidation number of nitrogen = +5

Oxidation number of oxygen = -2

Oxidation states of elements on product side:

Oxidation state of copper  = +2

Oxidation state of sulfur = -2

Oxidation number of silver  +1

Oxidation number of nitrogen = +5

Oxidation number of oxygen = -2

No change in oxidation states of elements is observed which means that the given reaction is not a redox reaction.

It is reaction of ion exchange. Substances react in the melt. As a result we receive alloy of four salts.

CuS + 2AgNO₃ ⇄ Cu(NO₃)₂ + Ag₂S

An atom is the smallest identifiable unit of a compound true or false

False that  atom  is the smallest  identifiable  unit  of  a compound.

The   smallest   identifiable  unit  of   a compound  is  the    Element.  Element  is the one which  make up the  compound  and element  is made up by atoms.  Example  of element  is  oxygen  and  hydrogen  which  make  up water (H2O) which is  a compound.

Hydrate name of CaCl2 * 8 H2O

Calcium chloride hydrate

Explain what each quantum number in a quantum number set tells you about the electron. Compare and contrast the locations and properties of two electrons within an atom that have the quantum number sets (2, 1, 0, +½) and (2, 0, 0, +½).

This is a bit long right now.

(n l m s) these are the numbers.

n=2, second orbital level.
l is the type of orbit: l=1 elongated, l=0 spherical
m=0, magnetic number: 0
s=spin, both have spin positive

You need still to round it up. Srry!

The first electron is in the n=2 shell, the p subshell (l=1), m sub l = 0 so it is in the middle space of its subshell's energy level diagram. It has a positive spin.

The second electron is in the n=2 shell, is in the s subshell (l=0), m sub l = 0 so it is in the middle space of its subshell's energy level diagram. It also has a positive spin.

A pure silver ring contains 5.15×1022 silver atoms. how many moles of silver atoms does it contain? express the number of moles to two significant figures.

1 mole ------------------ 6.02x10²³atoms
(moles) -----------------5.15x10²² atoms

moles silver = ( 5.15x10²²) * 1 / 6.02x10²³ =

moles silver = ( 5.15x10²²)  / 6.02x10²³ =

=> 0.085 moles of silver

(Select all that apply) All the reactants listed in the Table of Oxidation Potentials are:
*Oxidizing agents
*Reducing agents
*Oxidized by the reaction
*Reduced by the reaction

Reducing agents
Oxidized by the reaction

Sorry for late response on the answer.

All the reactants listed in the Table of Oxidation Potentials are reducing agents and oxidized by the reaction.