Question

Chlorine is used by textile manufacturers to bleach cloth. Excess chlorine is destroyed by its reaction with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), as follows. $$ \begin{aligned} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)+4 \mathrm{Cl}_{2}(g)+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \\ 2 \mathrm{NaHSO}_{4}(a q)+8 \mathrm{HCl}(a q) \end{aligned} $$ (a) How many moles of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) are needed to react with \(0.12 \mathrm{~mol}\) of \(\mathrm{Cl}_{2}\) ? (b) How many moles of \(\mathrm{HCl}\) can form from \(0.12 \mathrm{~mol}\) of \(\mathrm{Cl}_{2} ?\) (c) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) are required for the reaction $$ \text { of } 0.12 \mathrm{~mol} \text { of } \mathrm{Cl}_{2} ? $$ (d) How many moles of \(\mathrm{H}_{2} \mathrm{O}\) react if \(0.24 \mathrm{~mol} \mathrm{HCl}\) is formed?

Short answer

a) 0.03 mol Na2S2O3, b) 0.96 mol HCl, c) 0.60 mol H2O are required for the reaction with 0.12 mol Cl2; d) 0.15 mol H2O react if 0.24 mol HCl is formed.

Step-by-step solution

Determine moles of Na2S2O3 reacting with Cl2

Using stoichiometry based on the balanced chemical equation, determine the moles of sodium thiosulfate needed to react with the given moles of chlorine gas. The molar ratio between Cl2 and Na2S2O3 is 4:1.

Calculate moles of Na2S2O3 needed

Calculate the moles of Na2S2O3 needed. For every 4 moles of Cl2, 1 mole of Na2S2O3 is required. \[0.12 \text{ mol Cl2} \times \frac{1 \text{ mol Na2S2O3}}{4 \text{ mol Cl2}} = 0.03 \text{ mol Na2S2O3}\]

Determine moles of HCl forming from Cl2

To determine the moles of HCl that can form, look at the molar ratio between Cl2 and HCl in the balanced equation, which is 1:8.

Calculate moles of HCl formed

Calculate the moles of HCl that form from the given moles of Cl2. \[0.12 \text{ mol Cl2} \times \frac{8 \text{ mol HCl}}{1 \text{ mol Cl2}} = 0.96 \text{ mol HCl}\]

Determine moles of H2O reacting with Cl2

To determine the moles of H2O required for the reaction with Cl2, use the molar ratio from the chemical equation, which is Cl2:H2O = 1:5.

Calculate moles of H2O required

Calculate the moles of H2O required when 0.12 mol of Cl2 reacts. \[0.12 \text{ mol Cl2} \times \frac{5 \text{ mol H2O}}{1 \text{ mol Cl2}} = 0.60 \text{ mol H2O}\]

Determine moles of H2O reacting if 0.24 mol HCl is formed

To find the moles of H2O that react when 0.24 mol of HCl is formed, use the molar ratio between HCl and H2O in the balanced equation, which is 8:5.

Calculate moles of H2O reacting

Calculate the moles of H2O that would react when 0.24 mol of HCl is formed. Since for every 8 moles of HCl formed, 5 moles of H2O are required: \[0.24 \text{ mol HCl} \times \frac{5 \text{ mol H2O}}{8 \text{ mol HCl}} = 0.15 \text{ mol H2O}\]

Key concepts

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is foundational to understanding how substances interact in a chemical process. It involves calculations that relate the quantities of reactants and products in a chemical reaction based on the balanced chemical equation. For instance, when the equation for the sodium thiosulfate and chlorine reaction is balanced, it indicates that for every mole of sodium thiosulfate (Na2S2O3), four moles of chlorine (Cl2) are required, and eight moles of hydrochloric acid (HCl) alongside two moles of sodium bisulfate (NaHSO4) are produced.

Understanding stoichiometry is crucial as it helps predict how much of each reactant is needed and the amount of product that will be formed in a reaction. Let’s note some essential principles:

• The coefficients in a balanced chemical equation represent the molar ratio of the reactants and products involved in the reaction.
• The Law of Conservation of Mass dictates that the mass of the reactants must equal the mass of the products in a chemical reaction, meaning atoms are neither created nor destroyed.
• To carry out stoichiometric calculations, one must first ensure the chemical equation is balanced, reflecting the conservation of mass.

By applying these principles, we can solve various problems related to the quantities of substances in a reaction.

Sodium Thiosulfate and Chlorine Reaction

The reaction between sodium thiosulfate (Na2S2O3) and chlorine (Cl2) is an example of a redox reaction where chlorine is reduced and sodium thiosulfate is oxidized. In the textiles industry, this reaction plays a pivotal role in removing excess chlorine after the bleaching process to prevent the damage of fibers or fabrics. The balanced chemical equation for this reaction is:
\[\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(aq) + 4 \mathrm{Cl}_{2}(g) + 5\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaHSO}_{4}(aq) + 8 \mathrm{HCl}(aq)\]

In this equation, every mole of Na2S2O3 reacts with four moles of Cl2, producing eight moles of hydrochloric acid (HCl) and two moles of sodium bisulfate (NaHSO4), along with water. The stoichiometric coefficients tell us exactly how much of each substance is involved, which is critical when calculating the quantities required for a reaction to proceed completely.

Moles and Molar Ratio Calculations

The concept of moles and molar ratios is at the heart of stoichiometry. A mole is a unit that measures the amount of a substance. One mole contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) entities.

Molar ratios, derived from the balanced chemical equation, are used to convert between the moles of one substance to the moles of another. These ratios are essential in calculating the quantities needed for a complete reaction. To illustrate, consider the sodium thiosulfate and chlorine reaction:
\[\begin{aligned}&\text{Moles of Na2S2O3 needed:} \quad 0.12 \, \text{mol Cl}_{2} \times \frac{1 \, \text{mol Na2S2O3}}{4 \, \text{mol Cl}_{2}} = 0.03 \, \text{mol Na2S2O3}\&\text{Moles of HCl formed:} \quad 0.12 \, \text{mol Cl}_{2} \times \frac{8 \, \text{mol HCl}}{1 \, \text{mol Cl}_{2}} = 0.96 \, \text{mol HCl}\&\text{Moles of H2O required:} \quad 0.12 \, \text{mol Cl}_{2} \times \frac{5 \, \text{mol H2O}}{1 \, \text{mol Cl}_{2}} = 0.60 \, \text{mol H2O}\&\text{Moles of H2O reacting with formed HCl:} \quad 0.24 \, \text{mol HCl} \times \frac{5 \, \text{mol H2O}}{8 \, \text{mol HCl}} = 0.15 \, \text{mol H2O}\end{aligned}\]
When conducting stoichiometric calculations, it is critical to ensure the accuracy of molar ratios, as even slight deviations can lead to incorrect results. Therefore, understanding moles and molar ratios not only contributes to solving textbook exercises but is also key in practical chemistry, where the precise amount of substances are crucial for desired outcomes.