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Stoichiometry Based Problems Along with Concentration Terms - Definitions, Examples, Practice Problems & FAQs

Stoichiometry Based Problems Along with Concentration Terms - Definitions, Examples, Practice Problems & FAQs

Aishwarya and Shervin were playing in the garden on a Seesaw. Shervin was sitting on one end of the Seesaw. Can we say that the Seesaw is under a balanced state?

Well! The answer is no.

Now, to balance the Seesaw, Aishwarya sits on another side of the Seesaw and balances the swing.

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Similarly, chemical reactions need balancing. There is a need to balance the number of atoms on both sides of the reaction and this mainly gives information on which molar ratio reactants combine and forms a product. Once the chemical reaction is balanced, then only we can solve the problems based on the stoichiometry of the reaction.

So, let’s find out how to solve problems based on stoichiometry along with concentration terms.

Table of Contents

  • Concentration Terms
  • Mole-Mole Relationship in a Chemical Reaction
  • Stoichiometric Problems based on Concentration of Reagents
  • Practice Problems
  • Frequently Asked Questions - FAQs

Concentration Terms

There are different methods to represent the amount of solute present in a solution or concentration of the solution.

  • Mass by mass percentage
  • Mass by volume percentage
  • Volume by volume percentage
  • Mole fraction
  • Parts per million (ppm)
  • Molarity
  • Molality
  • Normality

Let’s discuss them all one by one.

Mass by mass percentage:

It is the amount of solute (in g) in 100 g of the solution.

mm%=mass of solute (g)mass of solution (g)×100

10 % mass by mass solution means 10 g of solute present in 100 g of the solution.

Mass by volume percentage:

It is the amount of solute (in g) in 100 mL of the solution.

mV%=mass of solute (g)volume of solution (mL)×100

10 % mass by volume solution means 10 g of solute present in 100 mL of the solution.

Volume by volume percentage:

It is the amount of solute (in mL) in 100 mL of the solution.

VV%=volume of solute (mL)volume of solution (mL)×100

10 % volume by volume solution means 10 mL of solute present in 100 mL of the solution.

Mole fraction:

It is the ratio of the number of moles of the solute or solvent and the total number of moles present in the solution. It is denoted by the symbol "".

Let, the number of moles of solute in a solution = n1

Number of moles of solvent in a solution = n2

Mole fraction of solute (solute)=n1n1 + n2

Mole fraction of solvent (solvent)=n2n1 + n2

Parts per million:

Parts per million can be defined as the amount of solute present in million particles of the solution.

It can be represented in terms of mass as:

Parts per million (ppm)  = mass of solute (g)mass of solution (g)×106

Molarity:

It is defined as the number of moles of solute present in 1 L of the solution.

Molarity (M)=No. of moles of solute (mol)Volume of the solution (L)=No. of moles of solute (mol)Volume of the solution (mL) × 1000

Molality:

It is defined as the number of moles of solute present in 1 kg of the solvent.

Molality (m)=No. of moles of solute (mol)Mass of the solvent (kg)=No. of moles of solute (mol)Mass of the solvent (kg) × 1000

Normality:

It is defined as the number of equivalents of the solute present in 1 L of the solution.

Normality (N)=No. of equivalents of solute (eq)Volume of the solution (L)=No. of equivalents of solute (eq)Volume of the solution (mL) × 1000

We need to understand the relation between the molarity and normality to solve stoichiometric problems.

⇒ Normality = Molarity × n-factor

Mole-Mole Relationship in a Chemical Reaction

Consider the reaction given below:

BF3+LiAlH4B2H6+LiF+AlF3

For mole-mole analysis a balanced chemical reaction is required.

4BF3+3LiAlH42B2H6+3LiF+3AlF3

In the above chemical equation, 4 moles of BF3 combines with 3 moles of LiAlH4 to form 2 moles of B2H6, 3 moles of LiF and 3 moles of AlF3.

Now, by using the above data, we can solve the stoichiometric problems. For example;

In the above reaction, how many moles of of AlF3 can be produced from 2 moles of BF3?

So, 4 moles of BF3 produces 3 moles of AlF3.

⇒ 1 mol of BF3 produces 34 mol of AlF3.

⇒ 2 mol of BF3 produces 34 × 2 mol of AlF3.

⇒ 2 mol of BF3 produces 1.5 mol of AlF3.

Stoichiometric Problems Based on Concentration of Reagents

We can take different examples to solve the stoichiometric problems.

Solving the stoichiometric problems when the molarity of the reactants and products are given:

For example;

Consider the following reaction:

H2SO4(aq)+NaOH (aq)  Na2SO4(aq)+H2O (l)

We need to find the molarity of 20 mL NaOH solution, when 50 mL of 2 molar H2SO4 solution reacted in the above reaction. Moreover, we also need to find the number of moles of Na2SO4 produced in the reactions.

The first step to solving the above problem would be to balance the reaction.

H2SO4(aq)+2NaOH (aq)  Na2SO4(aq)+2H2O (l)

Now, let's find out the number of moles of H2SO4 from the given data.

Number of moles of H2SO4=2 × 50 × 10-3 mol

From the balanced reaction;

1 mol of H2SO4 reacts with 2 mol of NaOH

⇒ 2 × 50 × 10-3 mol of H2SO4 reacts with 2× 2 × 50 × 10-3 mol of NaOH

⇒ 2 × 50 × 10-3 mol of H2SO4 reacts with 2 × 10-1 mol of NaOH

⇒ No. of moles of NaOH= 2 × 10-1 mol=0.2 mol

⇒ Molarity of NaOH solution =0.2 mol × 10320 mL= 10 mol L-1

Now, to find out the number of moles of Na2SO4 produced, we can use the stoichiometry of the reaction.

1 mol of H2SO4 produces 1 mol of Na2SO4

⇒ 2 × 50 × 10-3 mol of H2SO4 produces 1×2 × 50 × 10-3 mol of Na2SO4

⇒ 2 × 50 × 10-3 mol of H2SO4 produces 0.1 mol of Na2SO4

Solving the stoichiometric problems when the normality of the reactants and products are given:

For example;

Consider the same reaction:

H2SO4(aq)+NaOH (aq)  Na2SO4(aq)+H2O (l)

Now, we need to find out the normality of 20 mL NaOH solution, when 50 mL of 2 N H2SO4 solution reacted in the above reaction.

The first step to solving the above problem would be to balance the reaction.

H2SO4(aq)+2NaOH (aq)  Na2SO4(aq)+2H2O (l)

Now, let's find out the number of moles of H2SO4 from the given data.

Number of moles of H2SO4=N × Vn-factor 

n-factor of H2SO4= 2 (∵ it is furnishing 2 H+ ions per molecule in the reaction)

Number of moles of H2SO4=2 N × 50 mL × 10-32 

Number of moles of H2SO4=0.05 mol

From the balanced reaction;

1 mol of H2SO4 reacts with 2 mol of NaOH

⇒ 0.05 mol of H2SO4 reacts with 2 × 0.05 mol of NaOH

⇒ 0.05 mol of H2SO4 reacts with 0.1 mol of NaOH

⇒ No. of moles of NaOH=0.1 mol

⇒ Molarity of NaOH solution =0.1 mol × 10320 mL= 5 mol L-1=5 M

Now, Normality of NaOH = Molarity of NaOH × n-factor of NaOH

n-factor of NaOH=1 (∵ it is furnishing 1 OH- ion per molecule in the reaction)

⇒ Normality of NaOH = 1 × 5 =5 N

Practice problems

Q1. How much volume of 0.2 M H2SO4 is needed for the following reaction to yield 34 g of H2S?

8KI+5 H2SO44K2SO4+4I2+H2S+4H2O

  1. 35 L
  2. 45 L
  3. 50 L
  4. 25 L

Answer: D

Solution: Fron the given reaction, 8KI+5 H2SO44K2SO4+4I2+H2S+4H2O

Amount of H2S =34g (Given)

∴ Number of moles of H2S=Given weightMolar mass=34 g34 g mol-1=1 mol

We can observe that 5 moles of H2SO4 gives 1 mole of H2S (from the reaction)

We know;

Molarity (M)=No. of moles of solute (mol)Volume of the solution (L)

∴ Volume of H2SO4solutionMolarity=No. of moles of H2SO4

∴ Volume of H2SO4 solution=5 mol0.2 mol L-1=25 L

Hence, the correct option is (D).

Q2. What volume of 1.2 M Na2CO3 is required for the reaction

3Na2CO3(aq)+2AlCl3(aq)Al2(CO3)3(s)+6NaCl(aq) to occur with 0.30 M, 5 L AlCl3?

  1. 2.875 L
  2. 1.875 L
  3. 0.875 L
  4. 8.750 L

Answer: B

Solution:

Volume of AlCl3=5 L (Given)

Molarity of AlCl3=0.30 M (Given)

From here we can calculate the moles of AlCl3.

Moles of AlCl3= 5L0.30M=1.5 mol

From the chemical equation we can find mol-mol relationship:

3Na2CO3(aq)+2AlCl3(aq)Al2(CO3)3(s)+6NaCl(aq)

2 mol of AlCl3combines with 3 mol of Na2CO3

⇒ 1 mol of AlCl3combines with 32 mol of Na2CO3

∴ 1.5 mol of AlCl3 combines with 321.5=2.25 mol of Na2CO3

Molarity of Na2CO3=1.2 M (Given)

Number of moles of Na2CO3=2.25 mol (Calculated above)

Using this information we can calculate the volume of Na2CO3.

We know, Molarity (M)=No. of moles of solute (mol)Volume of the solution (L)

∴ Volume of Na2CO3=2.25 mol1.2 mol L-1=1.875 L

Hence correct option is B.

Q3. How much water is produced (in grams) when an excess of NaOH reacts with a 1 L solution of 3M HCl?

  1. 18 g
  2. 20 g
  3. 54 g
  4. 40 g

Answer: C

Solution: To explain the above-mentioned question we have to write the chemical equation for this.

HCl(aq)+NaOH(aq)H2O(l)+NaCl(aq)

We know that, Molarity (M)=No. of moles of solute (mol)Volume of the solution (L)

∴ Moles=Volume(L) Molarity

Moles of HCl=1 L3M

∴ Moles of HCl=3 mol

From the above equation we can see that 1 mol of HCl gives 1 mol of H2O.

∴ 3 mol of HCl produces 3 mol of water.

We know that 1 mol of H2O weighs 18 g;

∴ 3 mol of H2O weighs 183=54 g.

Hence the correct option is (C).

Q4. In the given reaction:

CaCO3(s)+2HCl(aq)CaCl2(s)+H2O(l)+CO2(g). What mass of CaCO3 is necessary for 0.75 M, 25 mL HCl to react?

  1. 93.75 g
  2. 0.9375 g
  3. 8375 g
  4. 0.8375 g

Answer: B

Solution: First calculate the moles when 25 mL of 0.75 M HCl is given.

Moles of HCl = 0.75M2510-3 L =0.01875 mol

From the given reaction, CaCO3(s)+2HCl(aq)CaCl2(s)+H2O(l)+CO2(g)

We have 2 mol of HCl reacts with 1 mol of CaCO3

1 mol of HCl reacts with 12 mol of CaCO3

⇒ 0.01875 mol of HCl reacts with 12×0.01875=0.009375 mol of CaCO3

For mass calculation, we can simply use, Number of moles=Given weightMolar Mass

Molar mass of CaCO3 comes out to be 100 g mol-1 .

∴ Mass of  CaCO3=0.009375mol×100g mol-1=0.9375 g

Hence, the correct option is (B).

Frequently Asked Questions-FAQs

Q. What is the basicity of an acid?
Answer:
The maximum number of H+ ions that can be furnished by a single molecule of acid is called the basicity of acid.

Acid

Basicity

HCl

1 (monobasic acid)

H2SO4

2 (dibasic acid)

H3PO4

3 (tribasic acid)

H3PO2

1 (monobasic acid)

Q. How are stoichiometry and moles related?
Answer:
In terms of moles, balanced chemical reactions are considered standard equations. Equivalences in moles are provided by a balanced chemical reaction, enabling stoichiometry calculations. A balanced chemical equation can be used to compute concentrations and bulk amounts of a single material expressed in moles.

Q. What do you understand about the law of mass action? How is it related to stoichiometry and concentration?
Answer:
According to the law of mass action, The rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants raised to their stoichiometric coefficient.

Let us say this is the mentioned reaction then the rate can be written as,

aA+bBcC+dD

Then, Rate ∝ [A]a [B]b

Q.Name the scientist that develops the concept of stoichiometry?
Answer:
German scientist Jeremias Richter is credited with the original discovery of stoichiometry. The term "stoichiometry," which has a difficult pronunciation and still confounds pupils, was created by Richter. Greek words for "element" and "metron," stoikheion, were used to create stoichiometry.

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