In this relationship, relate mass to volume. Mass is on the side where mass is provided and volume on the side
where volume is provided.
Note: 1 mole of a gas at S.T.P
= 22.4dm³(22400cm³)
Example 1: What volume of dry oxygen gas (measured at S.T.P) will be produced from the decomposition of
3.50g of potassium trioxochlorate(V)
(K = 39, O = 16, Cl = 35.5, 1 molar volume of gas at S.T.P = 22.4dm³)
Solution
$$\text{Equation of the reaction} $$
$${2KClO_3} \hspace{0.5em} ——> {2KCl} + {3O_2} $$
$$ {2KClO_3} \hspace{1em} ——> {3O_2} $$
$$ \text{2(39 + 35.5 + 16 × 3)} \hspace{0.2em} ——> 3 × 22.4 $$
$${245g} \hspace{1em} ——> {67.2dm³} $$
$${3.5g} \hspace{1em} ——> {x} $$
$$\text{cross multiply} $$
$$ {245x} = {3.5} × {67.2} $$
$${x} = \frac{235.2} {245} $$
$$ \text{The moles of } {O_2} \text{ produced = 0.96dm³} $$
Example 2: 4.2g of sodium reacted with HCl. Calculate the volume of hydrogen gas evolved at 15°C and
720mmHg if the saturated vapor pressure of water at 15°C is 30mmHg.
(Na = 23, H = 1, Cl = 35.5, 1 molar
volume of gas at S.T.P = 22.4dm³)
solution
To solve this, first find the volume of hydrogen gas at S T.P then find the volume of the dry gas
$${2Na} + {2HCl} \hspace{0.2em} ——> {2Nacl} + {H_2} $$
$$ {2Na} \hspace{1em}——> {H_2} $$
$$ {2 × 23g} \hspace{1em} ——> {1 × 22.4dm³} $$
$$ {46g} \hspace{1em} ——> {22.4dm³} $$
$$ {4.2g} \hspace{1em} ——> {x} $$
$$ {46x} \hspace{1em} ——> {22.4 × 4.2} $$
$$ {x} \hspace{1em}——> \frac{94.08}{46} $$
$$\text{Vol. of hydrogen at S.T.P ≈ 2dm³ } $$
To get the volume of dry hydrogen gas, use general gas equation;
$$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} $$
$${where} $$
$$ {P_1 = 760mmHg}$$
$${V_1 = 2dm³} $$
$$ {T_1 = 273k} $$
$$ {P_2 = 720 - 30 = 690mmHg} $$
$$ {V_2 = ?} $$
$$ { T_2 = 15 + 273 = 288K } $$
$$ {V_2} = \frac{760 × 2 × 288} {690 × 273} $$
$$\text{Vol. of dry hydrogen gas = 2.3dm³} $$
Mole to Volume Relationship
In this relationship, we relate mole and volume. Firstly relate the mole of the reaction to the general volume
of gas = 22.4dm³. Then relate what is given to get the volume or mole.
Example 1: What is the volume of 3.5 moles of oxygen gas (O2) at standard temperature and
pressure?
Solution
$$\text{1 mole of gas at S.T.P} = 22.4dm³ $$
$$ 1mole \hspace{1em} ———> \hspace{1em} 22.4dm³ $$
$$ \text{3.5 moles} \hspace{1em} ———> \hspace{1em} x $$
$$\text{volume} = {3.5 × 22.4} $$
$$\text{volume} = 78.4dm³ $$
Example 2: During a titration experiment, 0.05 moles of carbon (lV) oxide is liberated. What is the
volume of gas liberated?
Solution
$$\text{1 mole of gas at S.T.P} = 22.4dm³ $$
$$ 1mole \hspace{1em} ———> \hspace{1em} 22.4dm³ $$
$$ \text{0.05 moles} \hspace{1em} ———> \hspace{1em} x $$
$$\text{volume} = {0.05 × 22.4} $$
$$\text{volume} = 1.12dm³ $$
Example 3: calculate the maximum volume of NH3 gas at stp which could be obtained by
boiling 5dm³ of a 0.05M solution of ammonia (molar volume of stp=22.4dm³)?
Solution
$$\text{1 mole of gas at S.T.P} = 22.4dm³ $$
To get the moles of NH3 obtained on boiling,
$${n} = {concentration} × {volume} $$
$${n} = {0.05 × 5} $$
$${n} = \text{0.25 moles} $$
$$ 1mole \hspace{1em} ———> \hspace{1em} 22.4dm³ $$
$$ \text{0.25 moles} \hspace{1em} ———> \hspace{1em} x $$
$$\text{volume} = {0.25 × 22.4} $$
$$\text{volume} = 5.6dm³ $$
Example 4: Calculate the volume of 2M Hydrochloric acid which would be obtained by dissolving
560cm³ Hydrogen chloride gas (measured at s.t.p ) in water.
[molar volume of gas at s.t.p = 22.4dm³]?
Solution
$$\text{1 mole of gas at S.T.P} = 22.4dm³ $$
To get the moles of HCl gas at s.t.p
$$\text{560cm³} = 0.56dm³ $$
$$ 1mole \hspace{1em} ———> \hspace{1em} 22.4dm³ $$
$$ \text{x mole} \hspace{1em} ———> \hspace{1em} 0.56dm³ $$
$$\text{mole} = \frac{0.56}{22.4} $$
$$\text{moles of HCl gas} = \text{0.025 moles}$$
The volume of HCl acid in 0.025 moles of HCl gas can be determined by,
$${n} = {concentration} × {Volume} $$
$${Volume} = \frac {n}{concentration} $$
$${volume} = \frac{0.025}{2} $$
$${volume} = 0.0125dm³ $$
$${volume} = 12.5cm³ $$
