Problem 57 - Chemistry The Science in Context

Question

Write a balanced chemical equation for each of the following reactions: a. Dinitrogen pentoxide reacts with sodium metal to produce sodium nitrate and nitrogen dioxide. b. A mixture of nitric acid and nitrous acid is formed when water reacts with dinitrogen tetroxide. c. At high pressure, nitrogen monoxide decomposes to dinitrogen monoxide and nitrogen dioxide. d. Acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) burns and becomes carbon dioxide and water vapor.

Step-by-Step Solution

Answer

Answer: The balanced chemical equation for this reaction is: $$\mathrm{N}_{2}\mathrm{O}_{5} + 2\mathrm{Na} \rightarrow 2\mathrm{NaNO}_{3} + 2\mathrm{NO}_{2}$$

Step 1: Exercise a: Dinitrogen pentoxide reacts with sodium metal to produce sodium nitrate and nitrogen dioxide.

In order to write a balanced chemical equation, we first need to find the chemical formulas for each reactant and product: Dinitrogen pentoxide - \(\mathrm{N}_{2}\mathrm{O}_{5}\) Sodium metal - \(\mathrm{Na}\) Sodium nitrate - \(\mathrm{NaNO}_{3}\) Nitrogen dioxide - \(\mathrm{NO}_{2}\) Now, let's write the unbalanced equation: $$\mathrm{N}_{2}\mathrm{O}_{5} + \mathrm{Na} \rightarrow \mathrm{NaNO}_{3} + \mathrm{NO}_{2}$$ Next, we can balance the equation following these steps: 1. Balancing oxygen atoms: Since there are a total of 5 oxygen atoms on the reactant side and 4 on the product side, we multiply the sodium nitrate by 2 to get 6 oxygen atoms on each side making it 7 total. $$\mathrm{N}_{2}\mathrm{O}_{5} + \mathrm{Na} \rightarrow 2\mathrm{NaNO}_{3} + \mathrm{NO}_{2}$$ 2. Balancing sodium atoms: Since there are now 2 sodium atoms in the product side, we need to balance that by placing a 2 in front of the sodium metal in the reactant side: $$\mathrm{N}_{2}\mathrm{O}_{5} + 2\mathrm{Na} \rightarrow 2\mathrm{NaNO}_{3} + \mathrm{NO}_{2}$$ 3. Balancing nitrogen atoms: Lastly, balancing the nitrogen atoms by placing a 2 in front of the nitrogen dioxide in the product side: $$\mathrm{N}_{2}\mathrm{O}_{5} + 2\mathrm{Na} \rightarrow 2\mathrm{NaNO}_{3} + 2\mathrm{NO}_{2}$$ The balanced equation for this reaction is: $$\mathrm{N}_{2}\mathrm{O}_{5} + 2\mathrm{Na} \rightarrow 2\mathrm{NaNO}_{3} + 2\mathrm{NO}_{2}$$

Step 2: Exercise b: A mixture of nitric acid and nitrous acid is formed when water reacts with dinitrogen tetroxide.

First, let's find the chemical formulas for each reactant and product: Water - \(\mathrm{H}_{2}\mathrm{O}\) Dinitrogen tetroxide - \(\mathrm{N}_{2}\mathrm{O}_{4}\) Nitric acid - \(\mathrm{HNO}_{3}\) Nitrous acid - \(\mathrm{HNO}_{2}\) Now, let's write the unbalanced equation: $$\mathrm{H}_{2}\mathrm{O} + \mathrm{N}_{2}\mathrm{O}_{4} \rightarrow \mathrm{HNO}_{3} + \mathrm{HNO}_{2}$$ To balance the equation, we follow these steps: 1. Balancing hydrogen atoms: Multiply both nitric acid and nitrous acid by 2 to balance the hydrogen atoms: $$\mathrm{H}_{2}\mathrm{O} + \mathrm{N}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{HNO}_{3} + 2\mathrm{HNO}_{2}$$ 2. Balancing nitrogen atoms: There is no need to balance the nitrogen atoms, as they are already balanced. 3. Balancing oxygen atoms: Already balanced: The final balanced equation is: $$\mathrm{H}_{2}\mathrm{O} + \mathrm{N}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{HNO}_{3} + 2\mathrm{HNO}_{2}$$

Step 3: Exercise c: At high pressure, nitrogen monoxide decomposes to dinitrogen monoxide and nitrogen dioxide.:

For this exercise, the chemical formulas are: Nitrogen monoxide - \(\mathrm{NO}\) Dinitrogen monoxide - \(\mathrm{N}_{2}\mathrm{O}\) Nitrogen dioxide - \(\mathrm{NO}_{2}\) The unbalanced equation is: $$\mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O} + \mathrm{NO}_{2}$$ Now, let's balance the equation: 1. Balancing nitrogen atoms: Multiply the nitrogen monoxide by 3 to balance the nitrogen atoms: $$3\mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O} + \mathrm{NO}_{2}$$ 2. Balancing oxygen atoms: Already balanced. The final balanced equation is: $$3\mathrm{NO} \rightarrow \mathrm{N}_{2}\mathrm{O} + \mathrm{NO}_{2}$$

Step 4: Exercise d: Acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) burns and becomes carbon dioxide and water vapor.

First, let's find the chemical formulas for each reactant and product: Acetylene - \(\mathrm{C}_{2}\mathrm{H}_{2}\) Carbon dioxide - \(\mathrm{CO}_{2}\) Water vapor - \(\mathrm{H}_{2}\mathrm{O}\) Now, let's write the unbalanced equation: $$\mathrm{C}_{2}\mathrm{H}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$ To balance the equation, follow these steps: 1. Balancing carbon atoms: Since there are 2 carbon atoms in acetylene, multiply carbon dioxide by 2: $$\mathrm{C}_{2}\mathrm{H}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$ 2. Balancing hydrogen atoms: Multiply water vapor by 1 to balance the hydrogen atoms: $$\mathrm{C}_{2}\mathrm{H}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$ Lastly, we have the final balanced equation: $$\mathrm{C}_{2}\mathrm{H}_{2} + \mathrm{2.5} \mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}$$

Key Concepts

Chemical Reaction

A chemical reaction is a process that involves the transformation of one set of chemical substances to another. During this process, reactants undergo chemical changes to form new products, different in composition from the reactants. This is evident in exercises such as the reaction of dinitrogen pentoxide with sodium metal to produce sodium nitrate and nitrogen dioxide. A balanced chemical equation, like the ones we have in our exercises, provides a symbolic representation of the reaction, showing the substances involved and their relative amounts.

To make sense of these reactions, it's important to know the reactants and products, their chemical formulas, and how to balance these equations. Balancing an equation ensures that the number of atoms for each element is equal on both the reactant and product sides, adhering to the law of conservation of mass. The balanced reactions in the exercises show the transformations that occur on a molecular level in clear, precise terms.

Stoichiometry

The field of stoichiometry delves into the quantitative relationships between reactants and products in a chemical reaction. It is the mathematical backbone of chemistry, allowing scientists and students to calculate masses, volumes, and the number of particles involved in a reaction. Understanding stoichiometry is crucial for solving problems like the ones in the exercises, where balanced equations guide the calculation of required reactants or predicted products.

For instance, the reaction of acetylene with oxygen to form carbon dioxide and water requires us to apply stoichiometry to ensure that all reactants are used completely. The calculation of the exact amount of oxygen needed to react with a given quantity of acetylene is a typical stoichiometric calculation. Wonky stoichiometry could mean excess reactants or incomplete reactions, which in practical applications could be inefficient or even dangerous.

Law of Conservation of Mass

The law of conservation of mass is a fundamental concept in chemistry, stating that mass is neither created nor destroyed in a chemical reaction. This principle underpins the practice of balancing chemical equations, as represented in our textbook exercises. It ensures that the mass of the reactants equals the mass of the products. When you're balancing equations like the decomposition of nitrogen monoxide, you are essentially applying the law of conservation of mass.

For example, if we were to tally the mass of dinitrogen tetroxide and water before reacting, the sum should equal the mass of the nitric acid and nitrous acid after the reaction. Students learn to verify that the number of atoms for each element on both sides of the chemical equation is balanced, which indirectly confirms that the total mass remains unchanged. Grasping this law is critical for accurately predicting the amounts of substances produced or needed in any chemical process.