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${{m{K}}_2}{m{C}}{{m{r}}_2}{{m{O}}_7}\mathop  	o \limits^\Delta  {{m{K}}_2}{m{Cr}}{{m{O}}_4} + {{m{O}}_2} + X$. In the above reaction $X$ is

A.RAMA NAGESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Balancing Chemical Equations and Decomposition Reactions
	**Related Topics**: Stoichiometry, Redox Reactions, Properties of Chromates and Dichromates
	**Prerequisites**: Chemical Formulas, Balancing Chemical Equations
## SOLUTION PATHWAY
**Given Information**:
	Potassium dichromate (${m{K}}_2{m{Cr}}_2{m{O}}_7$) decomposes upon heating to produce potassium chromate (${m{K}}_2{m{CrO}}_4$), oxygen (${m{O}}_2$), and an unknown substance $X$.
**Method & Steps**:
	1. **Balance the chemical equation**
		Reasoning: The law of conservation of mass dictates that the number of atoms of each element must be the same on both sides of a balanced chemical equation.
		Formula/Rule:  Adjust coefficients of reactants and products to ensure equal numbers of each atom type on both sides.
	2. **Identify the unknown substance X**
		Working: Begin by balancing K and Cr atoms.  Since ${m{K}}_2{m{Cr}}_2{m{O}}_7$ contains 2 K atoms and 2 Cr atoms, and ${m{K}}_2{m{CrO}}_4$ also contains 2 K and 1 Cr, we might predict a coefficient of 2 for ${m{K}}_2{m{CrO}}_4$. However, this is not necessary as the question implies only one unknown product. 
${m{K}}_2{m{Cr}}_2{m{O}}_7 	o {m{K}}_2{m{CrO}}_4 + {m{O}}_2 + X$
We see 2 Cr on the left and 1 on the right in ${m{K}}_2{m{CrO}}_4$.  The remaining Cr must be part of X. Similarly, we have 7 O on the left, 4 in  ${m{K}}_2{m{CrO}}_4$, and 2 in ${m{O}}_2$, totaling 6. Thus, one O must be in X.  This leads to X as ${m{CrO}}_3$ initially. However, to balance Cr, X must contain 2 Cr. Therefore, X must be ${m{Cr}}_2{m{O}}_3$.
		Key Point: The balanced equation is $4{m{K}}_2{m{Cr}}_2{m{O}}_7 	o 4{m{K}}_2{m{CrO}}_4 + 2{m{Cr}}_2{m{O}}_3 + 3{m{O}}_2$. However, as the question implies only one molecule of $X$ is produced per molecule of ${m{K}}_2{m{Cr}}_2{m{O}}_7$ decomposed, we use the simplest ratio, which is $2{m{K}}_2{m{Cr}}_2{m{O}}_7 	o 2{m{K}}_2{m{CrO}}_4 + {m{Cr}}_2{m{O}}_3 + \frac{3}{2}{m{O}}_2$ or, equivalently,  ${m{K}}_2{m{Cr}}_2{m{O}}_7 	o {m{K}}_2{m{CrO}}_4 + 1/2 {m{Cr}}_2{m{O}}_3 + 3/4{m{O}}_2$.
	3. **Final Result**
		Result: $X = {m{Cr}}_2{m{O}}_3$
		Verification: The balanced equation confirms the identity of $X$.
## OPTION ANALYSIS
[A] :x: ${m{CrO}}_2$: Incorrect.  This would not balance the Cr atoms.
[B] ✓: ${m{Cr}}_2{m{O}}_3$: Correct. This balances the equation.
[C] :x: ${m{CrO}}_3$: Incorrect.  This would not balance the Cr atoms.
[D] :x: ${m{K}}_2{m{CrO}}_4$: Incorrect. This is already a product and would not balance the equation.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly balancing the equation: Double-check the number of atoms on both sides.
		2. Not considering the law of conservation of mass:  Remember that atoms cannot be created or destroyed in a chemical reaction.
	**Quick Tips**:
		1. Start by balancing the elements that appear in the fewest number of compounds.
		2. Double-check your work to ensure the equation is balanced.

A.RAMA NAGESWARARAO · Answer

CrO₂

A.RAMA NAGESWARARAO · Answer

The correct Option is "Cr₂O₃"

A.RAMA NAGESWARARAO · Answer

CrO₃

A.RAMA NAGESWARARAO · Answer

K₂CrO₄

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