Fill in the gaps to balance the equation.
Use the smallest set of whole numbers to balance the equation and include coefficients of "1" when appropriate.
[tex]$\square$ Na _3 N+$ $\square$ SrF _2 \rightarrow$ $\square$ NaF +$ $\square$ Sr _3 N_2[/tex]
Complete the table to determine how many atoms of each element are present in the reactants and products.
| Element | Reactant | Products |
| :------ | :------- | :------- |
| Na | [tex]$\square$[/tex] | [tex]$\square$[/tex] |
| N | [tex]$\square$[/tex] | [tex]$\square$[/tex] |
| Sr | [tex]$\square$[/tex] | [tex]$\square$[/tex] |
| F | [tex]$\square$[/tex] | [tex]$\square$[/tex] |
Answer (1)
Balance the chemical equation: 2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2 .
Count the number of atoms of each element on the reactant side.
Count the number of atoms of each element on the product side.
The balanced equation is 2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2 , with 6 Na, 2 N, 3 Sr, and 6 F atoms on both sides. The coefficients are 2 , 3 , 6 , 1 .
Explanation
Understanding the Problem We are given an unbalanced chemical equation and need to balance it using the smallest whole number coefficients. We also need to complete a table showing the number of atoms of each element on both sides of the balanced equation.
Balancing the Equation Let's start by balancing the equation: □ N a 3 N + □ S r F 2 → □ N a F + □ S r 3 N 2 . We will balance each element one by one.
Balancing Strontium (Sr) First, let's balance Strontium (Sr). There are 3 Sr atoms in S r 3 N 2 on the product side, so we need 3 S r F 2 on the reactant side. The equation becomes: □ N a 3 N + 3 S r F 2 → □ N a F + □ S r 3 N 2 .
Balancing Nitrogen (N) Next, let's balance Nitrogen (N). There are 2 N atoms in S r 3 N 2 on the product side, so we need 2 N a 3 N on the reactant side. The equation becomes: 2 N a 3 N + 3 S r F 2 → □ N a F + □ S r 3 N 2 .
Balancing Sodium (Na) Now the equation is: 2 N a 3 N + 3 S r F 2 → □ N a F + 1 S r 3 N 2 . Let's balance Sodium (Na). There are 2 * 3 = 6 Na atoms on the reactant side, so we need 6 N a F on the product side. The equation becomes: 2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2 .
Balancing Fluorine (F) Finally, let's balance Fluorine (F). There are 3 * 2 = 6 F atoms on the reactant side, and 6 F atoms in 6 N a F on the product side. The equation is now balanced: 2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2 .
Completing the Table Now, let's complete the table by counting the number of atoms of each element on both sides of the balanced equation.
Counting Atoms Na: Reactant = 2 * 3 = 6, Product = 6 * 1 = 6 N: Reactant = 2 * 1 = 2, Product = 1 * 2 = 2 Sr: Reactant = 3 * 1 = 3, Product = 1 * 3 = 3 F: Reactant = 3 * 2 = 6, Product = 6 * 1 = 6
Examples
Balancing chemical equations is essential in many real-world applications, such as in the pharmaceutical industry when synthesizing new drugs, in environmental science when studying pollution, and in materials science when creating new compounds. For instance, when designing a new drug, chemists need to ensure that the reaction produces the desired compound without any unwanted byproducts. By balancing the chemical equation, they can determine the exact amounts of reactants needed to achieve this goal, maximizing efficiency and minimizing waste. Similarly, in environmental science, understanding the balanced equation for a chemical reaction helps scientists predict the impact of pollutants on the environment and develop strategies to mitigate their effects. In materials science, balanced equations are crucial for creating new materials with specific properties, ensuring that the final product has the desired composition and structure.
