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\( Cu + HNO _ { 3 } \rightarrow Cu ( NO _ { 3 } ) _ { 2 } + NO + H _ { 2 } O \)

Ask by Frazier Harrington. in Nigeria
Jan 12,2025

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Answer

The balanced chemical equation is: \[ Cu + 4 HNO_3 \rightarrow Cu(NO_3)_2 + 2 NO + 2 H_2O \]

Solution

To balance the chemical equation \( Cu + HNO_3 \rightarrow Cu(NO_3)_2 + NO + H_2O \), we will follow these steps: 1. **Identify the number of atoms of each element on both sides of the equation.** 2. **Balance the equation by adjusting coefficients.** 3. **Ensure that the number of atoms for each element is the same on both sides.** ### Step 1: Count the Atoms - **Reactants:** - Cu: 1 - H: 1 - N: 1 - O: 3 - **Products:** - Cu: 1 (in \( Cu(NO_3)_2 \)) - H: 2 (in \( H_2O \)) - N: 2 (1 in \( Cu(NO_3)_2 \) and 1 in \( NO \)) - O: 7 (6 in \( Cu(NO_3)_2 \) and 1 in \( H_2O \)) ### Step 2: Balancing the Equation 1. **Balance Copper (Cu):** - Cu is already balanced (1 on each side). 2. **Balance Nitrogen (N):** - There are 2 N in the products (1 from \( Cu(NO_3)_2 \) and 1 from \( NO \)), so we need 2 \( HNO_3 \) in the reactants. - Update the equation: \[ Cu + 2 HNO_3 \rightarrow Cu(NO_3)_2 + NO + H_2O \] 3. **Balance Hydrogen (H):** - Now we have 2 H in the reactants (from \( 2 HNO_3 \)) and 2 H in the products (from \( H_2O \)), so H is balanced. 4. **Balance Oxygen (O):** - Reactants: \( 2 HNO_3 \) gives 6 O. - Products: \( Cu(NO_3)_2 \) gives 6 O and \( H_2O \) gives 1 O, totaling 7 O. - We need to adjust the number of \( H_2O \) to balance the O. We already have 6 O from \( Cu(NO_3)_2 \) and need 1 more from \( H_2O \), so we can keep it as is. ### Final Balanced Equation The balanced equation is: \[ Cu + 4 HNO_3 \rightarrow Cu(NO_3)_2 + 2 NO + 2 H_2O \] ### Summary The balanced chemical equation is: \[ Cu + 4 HNO_3 \rightarrow Cu(NO_3)_2 + 2 NO + 2 H_2O \] This ensures that the number of atoms for each element is the same on both sides of the equation.

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The Deep Dive

The reaction between copper and nitric acid is a classic demonstration of how metals interact with acids. In this case, copper (Cu) doesn’t react with dilute nitric acid but will react with concentrated nitric acid, leading to the formation of copper(II) nitrate and nitrogen monoxide, along with water. This showcases not just the properties of reagents but also the stoichiometry involved in such reactions — a valuable lesson in both chemistry and balancing equations. In practical terms, understanding this reaction is critical for metalworking and the production of metal nitrates, which are essential in fertilizers and explosives. By grasping how copper reacts to different concentrations of nitric acid, industries can formulate proper methods for dissolving or treating metals, avoiding hazardous situations and ensuring efficiency in production processes. Through this knowledge, one can appreciate the balance of chemistry and safety in real-world applications!

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