Question

If 45.98 g of sodium combines with an excess of chlorine gas to form 116.89 g of sodium chloride, what mass of chlorine gas is used in the reaction?

Short answer

The mass of chlorine gas used in the reaction is \(70.91 \text{ g}\).

Step-by-step solution

Calculate the mass of sodium chloride produced

We are given the mass of sodium chloride formed (116.89 g).

Calculate the difference between the mass of sodium chloride and the mass of sodium used

To get the mass of the chlorine used in the reaction, subtract the mass of sodium used (45.98 g) from the total mass of sodium chloride (116.89 g): Mass of chlorine = Mass of sodium chloride - Mass of sodium

Calculate the mass of chlorine used in the reaction

Using the formula from step 2, calculate the mass of chlorine: Mass of chlorine = 116.89 g (sodium chloride) - 45.98 g (sodium) Mass of chlorine = 70.91 g So, 70.91 g of chlorine gas was used in the reaction.

Key concepts

Law of Conservation of Mass

One of the fundamental principles in chemistry is the Law of Conservation of Mass, which states that the total mass of matter remains constant before and after a chemical reaction. This law implies that matter can neither be created nor destroyed in the course of a reaction.

In our exercise, the Law of Conservation of Mass is illustrated through the chemical reaction between sodium and chlorine gas to form sodium chloride. According to this principle, the mass of the reactants (sodium and chlorine gas) must equal the mass of the products (sodium chloride). Hence, if we start with 45.98 g of sodium and an excess of chlorine gas to produce 116.89 g of sodium chloride, the mass of chlorine gas reacted can be determined by the difference in mass, showcasing the law's validity.

Example:
Mass of sodium chloride (product) = Mass of sodium (reactant) + Mass of chlorine (reactant).
By rearranging this equation, we calculate the mass of chlorine needed in the reaction to validate the Law of Conservation of Mass.

Chemical Reaction Calculations

Chemical reaction calculations are essential for determining the quantities of reactants and products involved in a chemical reaction. These calculations often involve balancing chemical equations and using stoichiometry to find the relationships between reactants and products.

Stoichiometry is the tool that allows us to apply the Law of Conservation of Mass to individual atoms and molecules in a reaction. By knowing the amount of one reactant or product, we can calculate the amounts of others involved in the reaction. Our textbook example involves a simple stoichiometric calculation, where we find the mass of chlorine gas that combined with sodium to form sodium chloride.

Application:
With the previously calculated masses, we notice that the mass conservation principle is met when the mass of chlorine is subtracted from the mass of sodium chloride, aligning with the expected stoichiometric outcome. This is a straightforward example, but in more complex reactions, one may have to convert masses to moles or use molar ratios for accurate results.

Mole Concept

The mole concept is a fundamental aspect of stoichiometry that provides a bridge between the atomic realm and the macroscopic world we can measure. One mole represents Avogadro's number of particles (approximately \(6.022 \times 10^{23}\)), which is a fixed number of atoms or molecules, similar to a dozen meaning twelve.

In the context of our reaction, if we needed to calculate the number of moles of chlorine gas used, we would divide the mass of chlorine gas by its molar mass (the mass of one mole of chlorine gas). Knowing the number of moles could help in determining the number of particles and the proportion of reactants in balanced chemical equations, leading to more advanced calculations of reactant-product relationships in chemical reactions.

Illustrative Calculation:
For instance, after finding the mass of chlorine gas (70.91 g), we could convert it to moles by dividing it by the molar mass of chlorine gas (approximately 70.9 g/mole for \(Cl_2\)), concluding that approximately 1 mole of chlorine gas is used in the reaction.