Initially 3 moles of 'A' was taken in a 1 L container. The approx. moles of A left in the container when the following equilibrium estblished is `x xx 10^(-6)`. The value of 'x' is
`3A hArr B, K_(C)=8xx10^(15)`

Answer

To solve the problem, we need to determine the approximate moles of 'A' left in the container when the equilibrium is established for the reaction:

\[ 3A \rightleftharpoons B \]

Given that the equilibrium constant \( K_c = 8 \times 10^{15} \) is very high, we can infer that the reaction favors the formation of product 'B', meaning that most of the 'A' will have reacted to form 'B'.

### Step-by-step Solution:

1. **Initial Conditions**:
- We start with 3 moles of 'A' in a 1 L container. Therefore, the initial concentration of 'A' is:
\[ [A]_{initial} = 3 \, \text{mol/L} \]

2. **Change in Concentration**:
- Let \( y \) be the amount of 'A' that reacts to form 'B'. According to the stoichiometry of the reaction, for every 3 moles of 'A' that react, 1 mole of 'B' is formed. Hence, the change in concentration for 'A' will be:
\[ [A]_{change} = -3y \]
- The concentration of 'B' formed will be:
\[ [B]_{change} = y \]

3. **Equilibrium Concentrations**:
- At equilibrium, the concentration of 'A' will be:
\[ [A]_{equilibrium} = 3 - 3y \]
- The concentration of 'B' will be:
\[ [B]_{equilibrium} = y \]

4. **Equilibrium Constant Expression**:
- The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[B]}{[A]^3} \]
- Substituting the equilibrium concentrations into the expression:
\[ K_c = \frac{y}{(3 - 3y)^3} \]
- Given \( K_c = 8 \times 10^{15} \), we can set up the equation:
\[ 8 \times 10^{15} = \frac{y}{(3 - 3y)^3} \]

5. **Assumption for High \( K_c \)**:
- Since \( K_c \) is very large, we can assume that \( y \) is very close to 1 (meaning almost all of 'A' has reacted). Thus, we can approximate \( 3 - 3y \) as being very small.
- Let’s assume \( 3y \approx 3 \) (since most of 'A' has reacted):
\[ 3 - 3y \approx 0 \]

6. **Solving for \( y \)**:
- Rearranging the equation:
\[ y \approx 3 \times 10^{-5} \]
- Therefore, the moles of 'A' left at equilibrium will be:
\[ [A]_{equilibrium} = 3 - 3y \approx 3 - 3 \times 10^{-5} \approx 3 - 0.00003 \approx 2.99997 \text{ moles} \]
- The moles of 'A' left can be expressed in scientific notation as:
\[ [A]_{equilibrium} \approx 5 \times 10^{-6} \]

7. **Finding 'x'**:
- Since the problem states that the amount of 'A' left is \( x \times 10^{-6} \), we can conclude:
\[ x = 5 \]

### Final Answer:
The value of \( x \) is \( 5 \).