6 gm nitrogen on successive reaction with different compounds gets finally converted into 30 gm [`Cr(NH_(3))_(x)Br_(2)`] value of x is [ Atomic mass of Cr = 52, Br =80]

Answer

To solve the problem step by step, we need to find the value of \( x \) in the compound \( Cr(NH_3)_xBr_2 \) given that 6 grams of nitrogen is converted into 30 grams of this complex.

### Step 1: Calculate the number of moles of nitrogen
The atomic mass of nitrogen (N) is 14 g/mol. Therefore, the number of moles of nitrogen in 6 grams can be calculated using the formula:

\[ \text{Number of moles of N} = \frac{\text{mass}}{\text{molar mass}} = \frac{6 \, \text{g}}{14 \, \text{g/mol}} = \frac{3}{7} \, \text{mol} \]

### Step 2: Write the expression for the number of moles of the complex
The complex is given as \( Cr(NH_3)_xBr_2 \). The molar mass of the complex can be expressed as:

\[ \text{Molar mass of } Cr(NH_3)_xBr_2 = 52 + 17x + 80 \times 2 = 52 + 17x + 160 = 212 + 17x \, \text{g/mol} \]

### Step 3: Calculate the number of moles of the complex
The number of moles of the complex can be calculated using the total mass of the complex:

\[ \text{Number of moles of complex} = \frac{\text{mass}}{\text{molar mass}} = \frac{30 \, \text{g}}{212 + 17x \, \text{g/mol}} \]

### Step 4: Set up the equation relating moles of nitrogen and moles of the complex
From the stoichiometry of the reaction, we can relate the moles of nitrogen to the moles of the complex:

\[ \frac{3}{7} \cdot \frac{1}{x} = \frac{30}{212 + 17x} \]

### Step 5: Cross-multiply to solve for \( x \)
Cross-multiplying gives us:

\[ 3(212 + 17x) = 30 \cdot 7x \]

Expanding both sides:

\[ 636 + 51x = 210x \]

### Step 6: Rearranging the equation
Rearranging the equation to isolate \( x \):

\[ 636 = 210x - 51x \]
\[ 636 = 159x \]

### Step 7: Solve for \( x \)
Now, divide both sides by 159:

\[ x = \frac{636}{159} = 4 \]

### Conclusion
The value of \( x \) is 4. Therefore, the final formula for the complex is \( Cr(NH_3)_4Br_2 \).

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