Calculations based on Chemical Equations

A balanced chemical equation provides information regarding the relative amount, mass or volume of the reactants and products.

The fundamental behind all these relationships is the mole concept. The number of moles of any substance is directly related to its mass and its number of molecules or atoms. Therefore, different relationships can be written between the reactants and products. Some typical relationships are given alongside and below.

Some useful relationships are,

How to solve numerical problems based on chemical equations

The calculations based on chemical equations can be done as follows.

§  Write down the given amount, mass or volume of the substance, and identify the substance whose amount, mass or volume is to be found out.

§  Write down the balanced chemical equation.

§  Write down the amount, mass or volume of the concerned reactant and product, as indicated by the balanced chemical equation.

§  Calculate the unknown quantity by applying the unitary method.

Calculations based on mole-mole relationship

In a balanced chemical equation, the coefficients placed before any symbol or formula gives the number of moles of that substance involved in the reaction.

These coefficients therefore give the relative amounts (no. of moles) of the reactants and products. This relationship is illustrated through the following examples:

EXAMPLE 1.

A reaction proceeds in accordance with the chemical equation.

3CaCl₂        +      2K₃PO₄           ®     Ca₃(PO₄)₂    +      6KCl

calcium                  potassium               calcium                  potassium

chloride                  phosphate              phosphate              chloride

How many moles of potassium phosphate are needed to produce 0.076 moles of potassium chloride?

Solution: The balanced chemical equation is

                        3CaCl₂ + 2K₃PO₄ ® Ca₃(PO₄)₂ + 6KCl

                                     2 mol                      6 mol

This gives the relationship,

6 mol of KCl are obtained from 2 mol of K3PO4

1 mol of KCl is obtained from 2/6 mol of K3PO4

        0.076 mol of KCl is obtained from (2 × 0.076) / 6 mol of K3PO4

= 0.025 mol of K3PO4

Thus, to produce 0.076 mol of potassium chloride, 0.025 mol of potassium phosphate (K₃PO₄) is required.

Calculations based on mole-mass and mass-mass relationships

For making these calculations, the number of moles of the reactants and products are converted into mass by using their molar masses. Some typical calculations are illustrated below.

EXAMPLE 2.

Hydrogen reacts to form water as follows.

2H₂ + O₂ ® 2H₂O

How many grams of hydrogen are needed to react completely with 6.4 g of oxygen gas?                             (Atomic masses are : H = 1 u, O = 16 u.)

Solution: The chemical equation for the reaction, and the mass-mole relationship is,

2H₂            +      O₂            ®     2H₂O

                2 mol                 1 mol                 2 mol

        2 × (2 × 1) g       1 × (2 × 16) g              2 × (2 × 1 + 16) g

                4 g                    32 g                  36 g

Thus,

32 g of oxygen reacts completely with 4 g of hydrogen

1 g of oxygen reacts completely with (4 / 32) g of hydrogen

\ 6.4 g of oxygen reacts completely with (6.4 × 4 / 32) g = 0.8 g of hydrogen

Thus, 6.4 g of oxygen gas reacts completely with 0.8 g of hydrogen.

EXAMPLE 3.

Hydrazine (N₂H₄) and hydrogen peroxide (H₂O₂) are used together as a rocket fuel. The products of the reaction are nitrogen and water. How many grams of H₂O₂ are needed per kg (1000 g) of hydrazine carried by the rocket?

Solution: The reaction is,

Hydrazine + Hydrogen peroxide ® Nitrogen + Water

The balanced chemical equation is,

N₂H₄          +      2H₂O₂        ®     N₂     +      4H₂O

                1 mol                 2 mol

From the chemical equation of the reaction,

                1 mol of N₂H₄ = 2 mol of H₂O₂

or (2 × 14 + 4 × 1.0) g of N₂H₄ = 2 × (2 × 1.0 + 2 × 16.0) g of H₂O₂

                 (28 + 4) g of N₂H₄ = 2 × (34) g of H₂O₂

                        32 g of N₂H₄ = 68 g of H₂O₂

Thus, 32 g of N₂H₄ requires 68 g of H₂O₂

 1 g of N₂H₄ requires (68 / 32) g of H₂O₂

 1000 g of N₂H₄ requires (1000 × 68 / 32) g = 2125 g of H₂O₂

Therefore, mass of H₂O₂ required in the reaction is 2125 g, or 2.125 kg.

Calculations based on mass-volume and

volume-volume relationships

The following relationships are useful for solving problems based on mass-volume, and volume-volume relationships:

Molar volume of a gas at NTP = 22.4 L mol^–1                   (= 22400 mL)

 i.e., 1 mol of each gas under NTP conditions occupies a volume of 22.4 L.

These calculations are illustrated through the following examples:

EXAMPLE 4.

2.3 g of metallic sodium reacts with excess of water. Calculate the mass of sodium hydroxide formed. What is the volume of hydrogen evolved under NTP conditions?

Solution: Sodium reacts with water according to the equation,

        2Na(s)        +      2H₂O         ®     2NaOH(aq)         +      H₂(g)

        2 mol                 2 mol                                                 1 mol

        2 × 23 g                                     2 (23 + 16 + 1)            2 g

        46 g                                          80 g                          22.4 L (at NTP)

Thus, 2.3 g of sodium gives (2.3 × 80 / 46) = 4.0 g of NaOH

and 2.3 g of sodium gives (22.4 × 2.3 / 46) = 1.12 L of H₂ gas

Therefore,    Mass of NaOH formed = 4.0 g

        Volume of H₂ evolved at NTP = 1.12 L

EXAMPLE 5.

Calculate the volume of oxygen at NTP obtained by decomposing 12.26 g of KClO₃.

Solution: KClO₃ decomposes as follows,

                     2 mol                                                 3 mol

        2 × (39 + 35.5 + 48) g                                      3 × 22.4 L

        (2 × 122.5) g = 245g                                         67.2 L

Thus,

 12.26 g of KClO₃ gives (12.26 × 67.2 / 245) L = 3.36 L of oxygen

So, 12.26 g of KClO₃ on decomposing gives 3.36 L of oxygen gas at NTP.