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In the redox reaction, x KMnO(4) + NH(3)...

In the redox reaction, `x KMnO_(4) + NH_(3) rarr y KNO_(3) + MnO_(2) + MnO_(2) + KOH + H_(2)O`, x and y are

A

`x = 4, y = 6`

B

`x = 3, y = 8`

C

`x = 8, y = 6`

D

`x = 8, y = 3`

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The correct Answer is:
To solve the redox reaction \( x \text{KMnO}_4 + \text{NH}_3 \rightarrow y \text{KNO}_3 + \text{MnO}_2 + \text{KOH} + \text{H}_2\text{O} \) and find the values of \( x \) and \( y \), we will follow these steps: ### Step 1: Identify Oxidation States - In \( \text{KMnO}_4 \), manganese (Mn) has an oxidation state of +7. - In \( \text{MnO}_2 \), manganese has an oxidation state of +4. - In \( \text{NH}_3 \), nitrogen (N) has an oxidation state of -3. - In \( \text{KNO}_3 \), nitrogen has an oxidation state of +5. ### Step 2: Determine Changes in Oxidation States - For manganese: The change in oxidation state from +7 to +4 is a reduction of 3. - For nitrogen: The change in oxidation state from -3 to +5 is an oxidation of 8. ### Step 3: Set Up the Stoichiometric Coefficients To balance the changes in oxidation states, we can use the method of cross-multiplication: - Let \( x \) be the coefficient for \( \text{KMnO}_4 \) and \( y \) be the coefficient for \( \text{KNO}_3 \). - The ratio of the changes in oxidation states gives us: \[ \frac{x}{y} = \frac{8}{3} \] - This implies: \[ 8 \text{KMnO}_4 + 3 \text{NH}_3 \rightarrow 8 \text{KNO}_3 + \text{MnO}_2 + \text{KOH} + \text{H}_2\text{O} \] ### Step 4: Balance the Reaction Now we will balance the entire reaction: - For potassium (K), we have 8 from \( \text{KMnO}_4 \) and need 8 in the products. - For nitrogen (N), we have 3 from \( \text{NH}_3 \) and need 3 in \( \text{KNO}_3 \). - For oxygen (O), we will balance the remaining oxygen atoms from \( \text{H}_2\text{O} \). ### Step 5: Write the Final Balanced Equation After balancing all components, we arrive at the balanced equation: \[ 8 \text{KMnO}_4 + 3 \text{NH}_3 \rightarrow 3 \text{KNO}_3 + 4 \text{MnO}_2 + 5 \text{KOH} + 4 \text{H}_2\text{O} \] ### Step 6: Identify Values of \( x \) and \( y \) From the balanced equation: - \( x = 8 \) - \( y = 3 \) ### Final Answer Thus, the values of \( x \) and \( y \) are: - \( x = 8 \) - \( y = 3 \) ---

To solve the redox reaction \( x \text{KMnO}_4 + \text{NH}_3 \rightarrow y \text{KNO}_3 + \text{MnO}_2 + \text{KOH} + \text{H}_2\text{O} \) and find the values of \( x \) and \( y \), we will follow these steps: ### Step 1: Identify Oxidation States - In \( \text{KMnO}_4 \), manganese (Mn) has an oxidation state of +7. - In \( \text{MnO}_2 \), manganese has an oxidation state of +4. - In \( \text{NH}_3 \), nitrogen (N) has an oxidation state of -3. - In \( \text{KNO}_3 \), nitrogen has an oxidation state of +5. ...
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