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 | $\square$ | |
| N | $\square$ | $\square$ |
| Sr | $\square$ | $\square$ |
| F | $\square$ | $\square$ |
Answer (1)
Balance Sr atoms: 2 N a 3 N + 3 S r F 2 → □ N a F + □ S r 3 N 2
Balance N atoms: 2 N a 3 N + 3 S r F 2 → □ N a F + 1 S r 3 N 2
Balance Na atoms: 2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2
The balanced equation is: 2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2
Explanation
Understanding the Problem We are given an unbalanced chemical equation and asked 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 Strategy To balance the equation, we will follow a systematic approach, adjusting coefficients to ensure the number of atoms for each element is the same on both the reactant and product sides.
Balancing Strontium (Sr) Let's start by balancing the number of Strontium (Sr) atoms. We have 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 now looks like this:
□ N a 3 N + 3 S r F 2 → □ N a F + □ S r 3 N 2
Balancing Nitrogen (N) Next, we balance the number of Nitrogen (N) atoms. We have 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 + 1 S r 3 N 2
Balancing Sodium (Na) Now, let's balance the number of Sodium (Na) atoms. We have 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
Checking Fluorine (F) Finally, let's check if Fluorine (F) is balanced. We have 3 * 2 = 6 F atoms on the reactant side and 6 * 1 = 6 F atoms on the product side. So, Fluorine is already balanced.
The Balanced Equation The balanced equation is:
2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2
Now, let's complete the table to show 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
Final Answer
Element
Reactant
Products
Na
6
6
N
2
2
Sr
3
3
F
6
6
The balanced equation is:
2 N a 3 N + 3 S r F 2 → 6 N a F + 1 S r 3 N 2
Examples
Balancing chemical equations is essential in various real-world applications, such as in the pharmaceutical industry when synthesizing new drugs, in environmental science when analyzing pollution, and in materials science when creating new compounds. For example, when designing a new drug, chemists need to ensure that the reaction produces the desired compound in the correct amount, without any unwanted byproducts. This requires precise balancing of the chemical equation to determine the exact quantities of reactants needed. Similarly, in environmental science, balancing equations helps in understanding the stoichiometry of reactions that lead to pollution, allowing scientists to develop effective strategies for remediation. In materials science, balanced equations are crucial for synthesizing new materials with specific properties, ensuring that the elements combine in the correct ratios to achieve the desired outcome.
