Stoichiometry is concerned with the relationships between the quantities of reactants and products in chemical reactions. Kinetics id concerned with the rates at which reactions takes place. Because stoichiometry quantitatively relates a change in one reactant to the change in another when the reaction rate of one reactant become known, Stoichiometry may be used to determine the reaction rate of another in the reaction.

It is the measurement and is the name given to the quantitative relationships between the compounds in a chemical reaction. These quantitative relationships allow chemists to calculate the amounts of reactants needed for a reaction and to predict the quantity of product. Using stoichiometric methods chemists can determine the formulas of compounds and can simplify procedures in chemical analysis.

**Introduction to stoichiometry calculations:**

**2H**_{2 }+ O_{2} $\rightarrow$ 2H_{2}O

Stoichiometry is calculation of quantitative relationships of reactants and products.

In simpler terms it is measurement of ratio of quantities of reactants and the products participating in a reaction.

Whenever we write some reaction, it appears that the reactants and products may be participating by mass.

e.g.

It could be inferred that in order to have complete reaction without any residue, double the mass of hydrogen of that of oxygen would be needed.

However it is not so.

The quantitative relationship is not by masses but masses divided by molar masses.Thus it is nothing but by moles.

Hence in the above reaction, 2 moles and not grams of hydrogen would react with one mole of oxygen and not one gram of oxygen.

The reactants and products do so because the atomic masses are different for different elements. If the atomic numbers would have been the same, then all the elements would have entered into a reaction by ratio of their masses. But that is not the case. Since different elements have different atomic masses, the mole definition too is different, each one having different gram in a mole and hence the participation in a reaction is by moles.

### Stoichiometry Steps

To solve the problems based on chemical equations we need to keep the following points in mind.

- Write down balanced equation in molecular form.
- Underline those substances whose weights/volume have been mentioned in the numerical or whose weights/volumes have to be determined.
- Write the atomic weights/moles/molar volume of the underlined substances depending upon the nature of the problem.
- Write down the actual quantities of the substances which had been underlined. For those substances whose weights/volumes have been calculated just write the sign of interrogation (?).
- Calculate the result by unitary method.

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Some of the solved problems based on stoichiometry are given below. ### Solved Examples

**Question 1: **Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperature. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH_{3} is reacted with 90.4 g of CuO, which is the limiting reagent? How many grams of N_{2} will be formed?

** Solution: **
**Step 1:**

NH_{3} (g) + CuO (s) $\rightarrow$ Cu (s) + H_{2}O (g) + N_{2} (g)

Balancing the above reaction we get:

2NH_{3} (g) + 3CuO (s) $\rightarrow$ 3Cu (s) + 3H_{2}O (g) + N_{2} (g)

**Step 2:**

**Step 3:**

**Step 4:**

**Question 2: **How many liters of SO_{2} will be produced from 26.9 L O_{2}?

** Solution: **

**Step 1:**

**Step 2:**

**Step 3:**

**Step 4:**

**Step 5:**

**Question 3: **How many liters of H_{2} are created from the reaction of 20.0 g K?

** Solution: **

**Step1:**

**Step 2:**

**Step 3:**

**Step 4:**

**Step 5:**

**Question 4: **How many grams of Al can be created decomposing 9.8 g of Al_{2}O_{3}?

** Solution: **

**Step 1:**

**Step 2:**

**Step 3:**

**Step 4:**

**Step 5:**

**Question 5: **How many moles of HCl are needed to react with 0.87 moles of Al?

** Solution: **

**Step 1:**

**Step 2:**

**Step 3: **

Balance the equation

From the description above, we can translate the reaction into an equation as follows-

NH

Balancing the above reaction we get:

2NH

Compute the moles of the substances whose mass are given

Amount of NH_{3} given is 18.1g.

Amount of CuO given is 90.4 g.

1 mole of NH_{3 } $\rightarrow$ 17.03 g of NH_{3}

‘x’ moles of NH_{3 } $\rightarrow$ 18.1 g of NH_{3}

x = (18.1 g of NH_{3}) (1 mole of NH_{3}) / (17.03 g of NH_{3})

= 1.063 moles of NH_{3}

1 mole of CuO $\rightarrow$ 79.54 g of CuO

‘y’ mole of CuO $\rightarrow$ 90.4 g of CuO

y = (90.4 g of CuO) (1 mole of CuO) / (79.54 g of CuO)

= 1.14 moles of CuO

To determine the limiting reagent, use the mole ratio. Compare the ratio of the moles obtained from the amount given and from the balanced equation.

The mole ratio of 2NH_{3}: 3CuO = 2:3

OR

2 moles of NH_{3} react with 3 moles of CuO

1.06 moles of NH_{3} $\rightarrow$ ‘a’ moles of CuO

a = (1.06 moles of NH_{3}) (3 moles of CuO) / (2 moles of NH_{3})

= 1.59 moles of CuO

Thus, we require 1.59 moles of CuO to react with 1.06 moles of NH3. But we only have 1.14 moles of CuO. So, the amount of CuO is limiting.

So, the amount of the product formed will depend upon the amount of the limiting reactant.

CuO is the limiting reagent.

Use the amount or moles of the limiting reagent to determine the amount or moles of the unknown

The mole ratio of 3CuO: 1N_{2} = 3:1

3 moles of CuO $\rightarrow$ 1 mole of N_{2}

1.14 moles of CuO $\rightarrow$ ‘b’ moles of N_{2}

b = (1.14 moles of CuO) (1 mole of N_{2}) / (3 moles of CuO)

= 0.38 moles of N_{2}

1 mole of N_{2 } $\rightarrow$ 28 g of N_{2}

0.38 moles of N_{2 } $\rightarrow$ ‘c’ g of N_{2}

c = (0.38 moles of N_{2}) (28 g of N_{2}) / (1 mole of N_{2})

= 10.64 g of N_{2}

Balance the equation and calculate the ratios

S_{2} + O_{2} $\rightarrow$ SO_{2}

S_{2} +2 O_{2} $\rightarrow$ 2 SO_{2}

2O_{2}:1 S_{2} (2:1)

2O_{2}:2SO_{2} (1:1)

Find the volume of the given

26.9L of O_{2}

Calculate the moles of the given

1 mole of O_{2 } $\rightarrow$ 22.4 L of O_{2}

‘x’ mole of O_{2} $\rightarrow$ 26.9 L of O_{2}

x = (26.9 L of O_{2}) (1 mole of O_{2}) / (22.4 L of O_{2})

= 1.20 moles of O_{2}

Calculate the moles using the ratios

1 mole of O_{2 } $\rightarrow$ 1 mole of SO_{2}

1.20 moles of O_{2} $\rightarrow$ ’y’ moles of SO_{2}

y = (1.20 moles of O_{2}) (1 mole of SO_{2}) / (1 mole of O_{2})

= 1.2 moles of SO_{2}

Calculate the volume using the new moles

1 mole of SO_{2 } $\rightarrow$ 22.4 L of SO_{2}

1.20 moles of SO_{2 } $\rightarrow$ ‘z’ L of SO_{2}

z = (1.20 moles of SO_{2}) (22.4 L of SO_{2}) / (1 mole of SO_{2})

= 26.9 L of SO_{2}

Balance the equation and calculate the ratios

K + H_{2}O $\rightarrow$ KOH + H_{2}

2K + 2H_{2}O $\rightarrow$ 2KOH + H_{2}

2K: 2 H_{2}O (1:1)

2K: 2KOH (1:1)

2K: 1 H_{2} (2:1)

Find the mass of the given

20.0 g K are used in the reaction.

Calculate the moles of the given

1 mol of K $\rightarrow$ 39 g of K

‘y’ mol of K $\rightarrow$ 20 g of K

y = (20 g of K) (1 mol of K) / (39 g of K)

= 0.513 mol of K

Calculate the moles using the ratios

2 mol of K $\rightarrow$ 1 mol of H_{2}

0.513 mol of K $\rightarrow$ ‘z’ mol of H_{2}

z = (0.513 mol of K) (1 mol of H_{2}) / (2 mol of K)

= 0.256 mol of H_{2}

Calculate the volume using the new moles

1 mol of H_{2 } $\rightarrow$ 22.4 L of H_{2}

0.256 mol of H_{2 } $\rightarrow$ ‘a’ L of H_{2}

a = (0.256 mol of H_{2}) (22.4 L of H_{2}) / (1 mol of H_{2})

= 5.73 L of H_{2}

Balance the equation for the reaction and calculate the ratios

Al_{2}O_{3} $\rightarrow$ Al + O_{2}

2Al_{2}O_{3 }$\rightarrow$ 4 Al +3O_{2}

2Al_{2}O_{3}: 4Al (1:2) 2Al_{2}O_{3}: 3O_{2} (1:1.5)

Find the mass of the given

9.8 g Al_{2}O_{3} are decomposed.

Calculate the moles of the given

1 mole of Al_{2}O_{3 } $\rightarrow$ 102 g of Al_{2}O_{3}

So, ‘x’ moles of Al_{2}O_{3 } $\rightarrow$ 9.8 g of Al_{2}O_{3}

x = (9.8 g of Al_{2}O_{3}) (1 mole of Al_{2}O_{3}) / (102 g of Al_{2}O_{3})

= 0.096 moles of Al_{2}O_{3}

Calculate the moles using the ratios

1 mole of Al_{2}O_{3 } $\rightarrow$ 2 mol of Al

So, 0.096 moles of Al_{2}O_{3 } $\rightarrow$ ‘y’ mol of Al

y = (0.096 moles of Al_{2}O_{3}) (2 mol of Al)/ (1 mole of Al_{2}O_{3})

= 0.19 mol of Al

Calculate the mass using the new moles

1 mol of Al $\rightarrow$ 27 g of Al

So, 0.19 mol of Al $\rightarrow$ ‘z’ g of Al

z = (0.19 mol of Al) (27 g of Al) / (1 mol of Al)

= 5.1 g of Al

Final answer: 5.1 g of Al

Balance the equation and calculate the ratios

Al + HCl $\rightarrow$ AlCl_{3} + H_{2}

2Al + 6HCl $\rightarrow$ 2AlCl_{3} + 3H_{2}

2Al: 6HCl (1: 3)

2Al: 2 AlCl_{3} (1: 1)

2Al: 3H_{2} (1:1.5)

Find the moles of the given

0.87 moles of aluminum are reacted with hydrochloric acid.

Calculate the new moles using the ratios

Now, from the reaction, we know that every 2 moles of Al react with 6 moles of HCl

OR

Every 1 mole of Al reacts with 3 moles of HCl.

1 mole of Al → 3 moles of HCl

So, 0.87 moles of Al $\rightarrow$ ‘x’ moles of HCl

x = (0.87 moles of Al) (3 moles of HCl) / (1 mole of Al)

= 2.6 mol of HCl