Problem 103 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of fluorine is reacted with two mole of hot and concentrated KOH. The products formed are \(\mathrm{KF}, \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2} .\) The molar ratio of \(\mathrm{KF}, \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{O}_{2}\) respectively is (a) \(1: 2: 1\) (b) \(1: 2: 2\) (c) \(0.5: 1: 2\) (d) \(2: 1: 0.5\)

Step-by-Step Solution

Answer

The correct molar ratio is 2:1:0.5.

Step 1: Write the Chemical Equation

The chemical reaction between fluorine (F₂) and potassium hydroxide (KOH) is given as follows: \[ \mathrm{F_2} + \mathrm{KOH} \rightarrow \mathrm{KF} + \mathrm{O_2} + \mathrm{H_2O} \] The exact balancing of the equation will show us the molar ratio of the products.

Step 2: Balance the Chemical Equation

Start by balancing the number of moles for each element on both sides of the equation.1. Balance F: On the right, we have 2 F in \( \mathrm{KF} \), so we need 1 mole of \( \mathrm{F_2} \) on the left.2. Balance K: We have 1 mole of \( \mathrm{KF} \) on the right, so we need 1 mole of \( \mathrm{KOH} \) on the left, but the reaction requires 2 moles. Therefore, multiply all reactants and products accordingly.3. Balance O: On the right, we have 1 mole of \( \mathrm{O_2} \) and 1 mole of water \( \mathrm{H_2O} \), requiring 2 moles of \( \mathrm{KOH} \).After balancing, the equation becomes: \[ \mathrm{F_2} + 2\mathrm{KOH} \rightarrow 2\mathrm{KF} + \mathrm{O_2} + \mathrm{H_2O} \]

Step 3: Determine the Molar Ratios

From the balanced equation:- 1 mole of \( \mathrm{F_2} \) reacts with 2 moles of \( \mathrm{KOH} \) to produce 2 moles of \( \mathrm{KF} \), 1 mole of \( \mathrm{O_2} \), and 1 mole of \( \mathrm{H_2O} \).So, the molar ratio of \( \mathrm{KF}: \mathrm{H_2O}: \mathrm{O_2} \) is \( 2:1:1 \).

Key Concepts

Balancing Chemical Equations

Balancing a chemical equation is like solving a puzzle. You need to ensure that the number of each type of atom is the same on both sides of the equation. This follows the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. To balance an equation, you must adjust the coefficients (the numbers in front of compounds) but not the subscripts (the small numbers that are part of a chemical formula). In our original exercise, we started with the unbalanced equation: \[ \mathrm{F_2} + \mathrm{KOH} \rightarrow \mathrm{KF} + \mathrm{O_2} + \mathrm{H_2O} \] Firstly, focus on one element at a time. For instance, ensure that the number of fluorine atoms on both sides is the same by adjusting the coefficient in front of \( \mathrm{KF} \). Do the same for potassium, oxygen, and hydrogen. A helpful tip is to balance elements that appear in only one reactant and one product first, usually metals, before moving on to more complex molecules like \( \mathrm{O_2} \). After experimenting with the coefficients, the balanced version would look like this: \[ \mathrm{F_2} + 2\mathrm{KOH} \rightarrow 2\mathrm{KF} + \mathrm{O_2} + \mathrm{H_2O} \] Once balanced, the coefficients will show you how many moles of each substance are needed and produced in the reaction.

Molar Ratios

The molar ratio of a chemical equation provides insight into the proportional relationship between reactants and products. This is directly obtained from the balanced chemical equation's coefficients. A molar ratio is crucial because it helps you to calculate how much of a product you can make from a given amount of reactants. Consider the balanced equation again: \[ \mathrm{F_2} + 2\mathrm{KOH} \rightarrow 2\mathrm{KF} + \mathrm{O_2} + \mathrm{H_2O} \] This tells us that one mole of \( \mathrm{F_2} \) reacts with two moles of \( \mathrm{KOH} \) to produce two moles of \( \mathrm{KF} \), and both one mole each of \( \mathrm{O_2} \) and \( \mathrm{H_2O} \). Therefore, the molar ratio of \( \mathrm{KF}: \mathrm{H_2O}: \mathrm{O_2} \) is \( 2:1:1 \). Understanding these ratios is fundamental in predicting how much of each product you can expect from certain amounts of reactants, enabling cost-effective and efficient chemical production.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It involves calculations that use relationships found in chemical equations to deduce quantities such as mass, moles, and volumes associated with reactants and products. With a balanced equation like \[ \mathrm{F_2} + 2\mathrm{KOH} \rightarrow 2\mathrm{KF} + \mathrm{O_2} + \mathrm{H_2O} \] stoichiometry helps you determine the amount of each product formed from a certain amount of reactant using molar ratios. The steps for a basic stoichiometric calculation include:

• Convert all given information into moles (using the molecular weight if necessary).
• Use the balanced equation to set up conversion factors and equate moles of one substance to moles of others using molar ratios.
• Convert calculated moles of desired substance back to the required unit of measurement (grams, liters, etc.).

Successfully using stoichiometry allows chemists to predict and manage chemical reactions efficiently, ensuring all reactants are used optimally and wastage is minimized.