Stoichiometry of Reactions II
Stoichiometric Relationships

The stoichiometric relationship between reactants and products can be mass-mass, mass-volume, mole-volume, volume-volume, e.t.c.

Mass-Mass Relationship

In this relationship, we relate only the masses of the relating particles. The first step is to relate the molar masses then the mass given to get the unknown. This relationship is effective when mass of either product or reactant is given.
Note: Always ensure the equation is balanced before calculating and use the moles to multiply the molar mass

Example 1: Calculate the mass of magnesium oxide formed when 3g of magnesium is burnt in oxygen(Mg = 24, O = 16)

Solution

$$ \text{2Mg} + {O_2} \hspace{1em} ———> \text{2MgO} $$ $$\text{2Mg} \hspace{1em} ————> \text{2MgO} $$ $$\text{2 × 24g} \hspace{1em} ———> \text{2(24 + 16)} $$ $$\text{48g} \hspace{1em} ———> \text{80g} $$ $$\text{Therefore, 3g} \hspace{1em} ———> \text{x} $$ $$\text{cross multiply} $$ $$\text{48x} = \text{3 × 80} $$ $$\text{x} = \frac{240}{48} $$ $$\text{mass of magnesium oxide} = \text{5g} $$

Example 2: How many grams of HBr would exactly be required to react with 2g of ethyne? (C = 12, H = 1, Br = 80) (Jamb)

Solution

$$ \text{Equation of the reaction} $$ $${C_2H_2} + {2HBr} \hspace{1em} ——> {C_2H_4Br_2} $$ $$ {C_2H_2} \hspace{1em}——> {2HBr} $$ $$ \text{2 × 12 + 1 × 2} \hspace{1em}——> {2(1 + 80)} $$ $$\text{26g} \hspace{1em}———> \text{162g} $$ $$\text{:. 2g} \hspace{1em}———> x $$ $$\text{ cross multiply} $$ $$\text{26x} = \text{162 × 2}$$ $$\text{x} = \frac{324}{26} $$ $$\text{mass of HBr} = \text{12.46g} $$

Example 3: Calculate the mass of copper deposited when 6.5g of granulated zinc reacts with excess copper (II) tetraoxosulphate(VI) according to the equation
Zn + CuSO4 ———> ZnSO4 + Cu
[Cu = 64, Zn = 65] (Waec)

Solution

$$\text{Zn} \hspace{1em} ———> \text{Cu} $$ $$\text{65g} \hspace{1em} ———> \text{64g} $$ $$\text{6.5g} \hspace{1em} ———> \text{x} $$ $$\text{cross multiply} $$ $$\text{65x} = \text{6.5 × 64} $$ $$\text{x} = \frac{416}{65} $$ $$\text{the mass of copper deposited} = \text{6.4g} $$

Mole to mole relationship

This relationship exists when the moles of the reactants or products is given. Here, relate the moles of the relating particles in the balanced equation first before relating with the number of moles given.

Example 1: Given the combustion of reaction of butane as
2C4H10 + 13O2 ——> 8CO2 + 10H2O
Find the moles of CO2 produced in the complete combustion of 5.2 moles of butane

Solution

$$ {2C_4H_{10}} \hspace{1em} ——> {8CO_2} $$ $$ \text{2 moles} \hspace{1em} ——> \text{8 moles} $$ $$ \text{5.2 moles} \hspace{1em} ——> \text{x} $$ $$ \text {cross multiply} $$ $$ \text{2x} = \text{5.2 × 8} $$ $$ \text{x} = \frac{41.6}{2} $$ $$ \text{Number of moles of } {CO_2} = \text{20.8 moles } $$

Example 2: How many moles of water is obtained when 5 moles of hydrogen is sparked with oxygen

Solution

$$\text{Equation of reaction} $$ $${2H_2} + {O_2} \hspace{0.5em}——> {2H_2O} $$ $$\text{2 mole of }{H_2} \hspace{0.5em} ——> \text{2 moles of }{H_2O} $$ $$ \text{5 moles} \hspace{1em} ———> {x} $$ $$ \text{cross multiply} $$ $$ \text{2x} = \text{5 × 2} $$ $$ \text{x} = \frac{10} {2} $$ $$\text{The moles of water = 5 moles} $$

Mass to Mole Relationships

In this relationship, relate mass and the number of moles. Mass on the side for which mass was given and moles on the side for which mass was not given.
Here 1 mole of the relating particle is equivalent to its molar mass.

Example 1: What is the number of moles of HNO3 required to dissolve 20.5g of limestone.

Solution

Equation of reaction
CaCO3 + 2HNO3 ———> Ca(NO3)2 + H2O + CO2 $${CaCO_3} \hspace{1em} ——> 2{HNO_3} $$ $${100g} \hspace {1em} ——> \text{2 moles} $$ $${20.5g} \hspace{1em}——> x $$ $$\text{cross multiply} $$ $$\text{100x} = \text{20.5 × 2} $$ $${x} = \frac{41} {100} $$ $$\text{ number of moles = 0.41mole} $$

Example 2: Calculate mass of carbon(IV) oxide produced from the combustion of 2 moles of propane according to the equation
C3H8 + 5O2 ——> 3CO2 + 4H2O

Solution

$$ {C_3H_8} \hspace{1em} ——> {CO_2} $$ $$ \text { 1 mole} \hspace{1em} ——> \text{3(12 + 16 × 2) } $$ $$ \text { 1 mole} \hspace{1em} ——> {132g} $$ $$\text{2 moles} \hspace{1em} ——> {x} $$ $${x} = \text{132 × 2} $$ $$ \text{ the mass of } {CO_2} \text{ is = 264g} $$