Problem 51 - Chemistry The Science in Context

Question

Fluorine is a very reactive, dangerous element, as can be seen in the following unbalanced equation: $$\mathrm{F}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HF}(a q)+\mathrm{O}_{2}(g)$$ After balancing this chemical equation, what is the coefficient of HF?

Step-by-Step Solution

Answer

Answer: The coefficient of HF in the balanced chemical equation is 2.

Step 1: Write the unbalanced equation

Begin by writing down the given unbalanced chemical equation: $$\mathrm{F}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HF}(a q)+\mathrm{O}_{2}(g)$$

Step 2: Balancing hydrogen atoms

Count the number of hydrogen atoms on both sides of the equation. On the left side, we have 2 hydrogen atoms in one \(\mathrm{H}_{2} \mathrm{O}\) molecule, and on the right side, we have 1 hydrogen atom in one \(\mathrm{HF}\) molecule. To balance these, we need to multiply the \(\mathrm{HF}\) by 2: $$\mathrm{F}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2\mathrm{HF}(a q)+\mathrm{O}_{2}(g)$$

Step 3: Balancing fluorine atoms

Now, we count the number of fluorine atoms on both sides. On the left side, we have 2 fluorine atoms in one \(\mathrm{F}_{2}\) molecule, and on the right side, we have 2 fluorine atoms in two \(\mathrm{HF}\) molecules. They are already balanced.

Step 4: Balancing oxygen atoms

Lastly, we must balance the oxygen atoms. On the left side, we have 1 oxygen atom in one \(\mathrm{H}_{2} \mathrm{O}\) molecule, and on the right side, we have 2 oxygen atoms in one \(\mathrm{O}_{2}\) molecule. To balance the oxygen atoms, we need to multiply the \(\mathrm{H}_{2} \mathrm{O}\) by 2: $$\mathrm{F}_{2}(g)+2\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2\mathrm{HF}(a q)+\mathrm{O}_{2}(g)$$ The chemical equation is now balanced. The coefficient of \(\mathrm{HF}\) in the balanced equation is 2.

Key Concepts

Chemical Reaction

A chemical reaction is the process through which one or more substances, known as reactants, are converted into one or more different substances, called products. This transformation occurs as a result of the rearrangement of atoms and the breaking and forming of chemical bonds.

In the provided exercise, F₂ reacts with H₂O to produce HF and O₂. During this reaction, fluorine (F₂) and water (H₂O) are the reactants, and hydrogen fluoride (HF) and oxygen (O₂) are the products. The reaction showcases the transformation where fluorine, being highly reactive, interacts with water to yield hydrogen fluoride and oxygen gas. It's important to understand that chemical reactions follow the Law of Conservation of Mass, meaning that matter is neither created nor destroyed during a reaction, which is why we balance chemical equations to reflect this law.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantification of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a reaction.

The concept of stoichiometry is grounded on the balanced chemical equation, from which the stoichiometric coefficients indicate the proportional amounts of each substance involved. In our example, after balancing, for every 1 molecule of F₂ and 2 molecules of H₂O, we produce 2 molecules of HF and 1 molecule of O₂. Stoichiometry is essential for calculating yields, determining limiting reagents, and will help in understanding how much product is formed, given a certain amount of reactants.

Chemical Equation Balancing

Balancing a chemical equation is a necessary step to accurately reflect the Law of Conservation of Mass. It ensures that the same number of atoms of each element are present on both sides of the equation.

To achieve this, we adjust coefficients, which are the numbers placed in front of molecules, without altering the chemical identity of the substances involved. In our exercise, we begin with unbalanced reactants and products and progress through a series of steps to balance the equation:

• First, we balanced hydrogen by ensuring both sides of the equation had an equal number of hydrogen atoms.
• Next, we confirmed that the fluorine atoms were already balanced.
• Last, we balanced the oxygen atoms, which required altering the coefficient of water (H₂O) on the reactant side to match the oxygen output on the product side.

Through these steps, we achieved a balanced chemical equation that obeys the Law of Conservation of Mass.

Reactivity of Fluorine

Fluorine, F₂, is known to be one of the most reactive elements on the periodic table. Its high reactivity is due to its electronegativity and its desire to gain one more electron to achieve a noble gas configuration.

This reactivity is on full display in our exercise, where fluorine gas reacts vigorously with water, splitting the H₂O molecules and forming HF and O₂. The reactivity of fluorine also implies that it can react under conditions where other elements might not, and it forms very strong bonds with other elements due to its ability to draw electrons towards itself. The understanding of fluorine's reactivity is crucial, especially in handling and storage, as it can be dangerous and can react with a wide array of substances, sometimes explosively.