Question

Sodium hypochlorite, the primary component in commercial bleach, is prepared by bubbling chlorine gas through solutions of sodium hydroxide. $$\mathrm{NaOH}(a q)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaOCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ How many moles of sodium hypochlorite can be prepared from \(32.5 \mathrm{~g}\) of \(\mathrm{NaOH}\) ?

Short answer

0.8125 moles of NaOCl can be prepared.

Step-by-step solution

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is \[ \mathrm{NaOH}(aq) + \mathrm{Cl}_2(g) \longrightarrow \mathrm{NaOCl}(aq) + \mathrm{H}_2\mathrm{O}(l) \]. This equation indicates that 1 mole of \( \mathrm{NaOH} \) reacts to produce 1 mole of \( \mathrm{NaOCl} \).

Calculate Molar Mass of NaOH

To find the molar mass of \( \mathrm{NaOH} \), add the atomic masses of sodium (Na), oxygen (O), and hydrogen (H): \( 22.99 \ g/mol + 16.00 \ g/mol + 1.01 \ g/mol = 40.00 \ g/mol \).

Convert Grams of NaOH to Moles

Use the molar mass from Step 2 to convert the given mass of \( \mathrm{NaOH} \) to moles. \[ \text{Moles of NaOH} = \frac{32.5 \ g}{40.00 \ g/mol} = 0.8125 \ mol \].

Calculate Moles of NaOCl Produced

According to the balanced equation, 1 mole of \( \mathrm{NaOH} \) produces 1 mole of \( \mathrm{NaOCl} \). Therefore, the moles of \( \mathrm{NaOCl} \) produced will be equal to the moles of \( \mathrm{NaOH} \). \[ 0.8125 \ mol \text{ of } \mathrm{NaOH} \rightarrow 0.8125 \ mol \text{ of } \mathrm{NaOCl} \].

Key concepts

Chemical Reactions

In the realm of chemistry, a chemical reaction is a process that leads to the transformation of one set of chemical substances to another. This means that the materials you start with, known as reactants, turn into completely different substances, known as products. The process always involves the breaking and forming of chemical bonds.

In the exercise with sodium hypochlorite, we are dealing with a double displacement type of chemical reaction. Here, the chlorine gas reacts with sodium hydroxide solution to produce sodium hypochlorite and water. A chemical equation not only represents this reaction symbolically but also shows the conservation of mass and the stoichiometric relationships between the reactants and products.

A balanced chemical equation is crucial as it ensures that the same number of each type of atom appears in both the reactants and the products. In our equation:

• 1 molecule of NaOH reacts with 1 molecule of Cl₂
• to form 1 molecule of NaOCl and 1 molecule of H₂O

This indicates a one-to-one-to-one-to-one ratio, guiding us in stoichiometric calculations.

Mole Concept

The mole concept is fundamental to understanding chemical reactions. It provides a bridge between atoms/molecules and grams. One mole is defined as exactly 6.022 x 10²³ particles, which is Avogadro's number.

This unit allows chemists to count particles in substances by relating the amount of substance to its mass. In the sodium hypochlorite exercise, the word 'mole' helps us measure the reactants and products accurately. This leads to precise stoichiometric calculations.

To convert grams to moles and vice versa, one must know the molar mass (mass of one mole) of a substance. For sodium hydroxide (NaOH), this is calculated by adding the atomic masses of its constituent atoms:

• Sodium (Na): 22.99 g/mol
• Oxygen (O): 16.00 g/mol
• Hydrogen (H): 1.01 g/mol

Thus, the molar mass of NaOH is 40.00 g/mol. Knowing this allows the conversion from grams to moles, which is key to determining recipe-like amounts needed for chemical reactions.

Mass-to-Mole Calculations

Mass-to-mole calculations are essential in chemistry, as they allow conversion between the mass of a substance and the amount of molecules or particles contained within that mass. This type of calculation is fundamental when dealing with chemical equations, as it specifically uses the molar mass of substances.

In the exercise example with sodium hypochlorite, we are given the mass of sodium hydroxide (NaOH) and need to find how many moles of it are present. The formula needed is:\[\text{Moles of } \mathrm{NaOH} = \frac{\text{Mass of } \mathrm{NaOH} (\text{g})}{\text{Molar Mass of } \mathrm{NaOH} (\text{g/mol})}\]By substituting the given mass of NaOH (32.5 g) and its molar mass (40.00 g/mol) into this equation, we find that 0.8125 moles of NaOH are present.

This straightforward calculation, derived from grams, is a stepping stone to figuring out the extent of a reaction and involves a direct connection to the number of moles of other substances, such as sodium hypochlorite in the end.