Question

Challenge For each of the following, balance the chemical equation; interpret the equation in terms of particles, moles, and mass; and show that the law of conservation of mass is observed. a) ____\(\mathrm{Na}(\mathrm{s})+\)____\(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow\) ____\(\mathrm{NaOH}(\mathrm{aq})+\) ____\(\mathrm{H}_{2}(\mathrm{g})\) b) ____\(Z n(s)+\) ____\(\mathrm{HNO}_{3}(\mathrm{aq}) \rightarrow\) ____\(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+\) ____\(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

Short answer

For reaction a, the balanced equation is \(2 \text{Na(s)} + 2 \text{H}_{2}\text{O(l)} \rightarrow 2 \text{NaOH(aq)} + \text{H}_{2}\text{(g)}\), where 2 moles of Na react with 2 moles of H2O to produce 2 moles of NaOH and 1 mole of H2. The law of conservation of mass is observed, as the masses of reactants and products are equal. For reaction b, the balanced equation is \( \text{Zn(s)} + 2\text{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+ \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \), where 1 mole of Zn reacts with 2 moles of HNO3 to produce 1 mole of Zn(NO3)2 and 1 mole of H2O. The law of conservation of mass is also observed in this reaction, as the masses of reactants and products are equal.

Step-by-step solution

Balance the equation

To balance the equation, we should make sure the same number of each element's atoms are present on both the reactants and products side. Balanced equation: \(2 \text{Na(s)} + 2 \text{H}_{2}\text{O(l)} \rightarrow 2 \text{NaOH(aq)} + \text{H}_{2}\text{(g)}\) Step 2: Interpreting the balanced equation in terms of particles, moles, and mass

Interpretation in terms of particles, moles, and mass

From the balanced equation, we know that \(2 \, \text{moles}\) of Na react with \(2 \, \text{moles}\) of H2O to produce \(2 \, \text{moles}\) of NaOH and \(1 \, \text{mole}\) of H2. Step 3: Showing the Law of Conservation of Mass

Law of Conservation of Mass

To show the conservation of mass, we can add up the masses of reactants and products. The sum of the masses of reactants should equal the sum of the masses of products. Mass of Reactants = \(2 \times \text{M(Na)} + 2 \times \text{M(H}_{2}\text{O)}\) Mass of Products = \(2 \times \text{M(NaOH)} + \text{M(H}_{2})\) Since the masses of reactants and products are equal, the law of conservation of mass is observed in this reaction. #Equation_b#: Given chemical equation for reaction b: Zn(s) + HNO3(aq) -> Zn(NO3)2(aq) + H2O(l) Step 1: Balancing the chemical equation

Balance the equation

To balance the equation, we should make sure the same number of each element's atoms are present on both the reactants and products side. Balanced equation: \( \text{Zn(s)} + 2\text{HNO}_{3}(\mathrm{aq}) \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+ \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \) Step 2: Interpreting the balanced equation in terms of particles, moles, and mass

Interpretation in terms of particles, moles, and mass

From the balanced equation, we know that \(1 \, \text{mole}\) of Zn reacts with \(2 \, \text{moles}\) of HNO3 to produce \(1 \, \text{mole}\) of Zn(NO3)2 and \(1 \, \text{mole}\) of H2O. Step 3: Showing the Law of Conservation of Mass

Law of Conservation of Mass

To show the conservation of mass, we can add up the masses of reactants and products. The sum of the masses of reactants should equal the sum of the masses of products. Mass of Reactants = \(\text{M(Zn)} + 2 \times \text{M(HNO}_{3})\) Mass of Products = \(\text{M(Zn(NO}_{3})_{2}) + \text{M(H}_{2}\text{O)}\) Since the masses of reactants and products are equal, the law of conservation of mass is observed in this reaction.

Key concepts

Law of Conservation of Mass

Understanding the Law of Conservation of Mass is crucial when it comes to balancing chemical equations. This fundamental principle states that in a closed system, mass cannot be created or destroyed during a chemical reaction. Instead, mass is conserved; the total mass of the reactants must equal the total mass of the products. In the context of our exercise with sodium (Na) reacting with water (H2O), we've seen in the solution that by carefully balancing the equation, the mass on both sides remains equal, effectively demonstrating this law.

When students balance the chemical equation, they practice applying this law by ensuring that the number of atoms for each element is the same on both sides of the equation. This not only validates the theory but also emphasizes the importance of precision in chemical calculations.

Chemical Reactions

Chemical reactions are processes in which reactants transform into products through the breaking and formation of chemical bonds. These reactions are depicted through chemical equations, which summarize the substances involved and their phase of matter (solid, liquid, gas, aqueous). The task of balancing these equations is not only a mathematical exercise but also a reflection of the actual physical process. When balancing the sodium and zinc reactions, notice how the reactants interact and the products that form. By analyzing the balanced equations, one can predict the outcome of the reaction—whether it's the formation of sodium hydroxide and hydrogen gas or zinc nitrate and water.

Stoichiometry

Stoichiometry is a section of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Through stoichiometry, you can calculate the amount of reactants required to produce a certain amount of product, or vice versa. It relies on the balanced chemical equation to provide the ratios of the molecules, generally expressed in moles. For example, in the balanced sodium and water reaction, the stoichiometry reflects a 1:1:1:0.5 ratio between Na, H2O, NaOH, and H2, respectively. Understanding stoichiometry is key to predicting the yield of a reaction and is a foundational concept for anyone studying chemistry.

Mole Concept

The mole concept is a bridge between the microscopic world of atoms and the macroscopic world we interact with every day. It allows chemists to count atoms, molecules, and ions in bulk quantities by using the mole unit, which is defined as the amount of substance containing as many entities as there are atoms in 12 grams of isotopically pure carbon-12. This concept makes it possible to talk about the chemical quantity in terms of mass, which can then be measured and manipulated in the lab. In the balanced equations provided, the mole concept helps to interpret how many moles of each substance react or are produced—and when plugged into stoichiometry, this concept becomes a powerful tool for quantitative analysis in chemistry.