The set of numerical coefficients that balances the chemical
equation
`K_(2)CrO_(4)+HCl to K_(2)Cr_(2)O_(7)+KCl+H_(2)O`

Options

A. 1,1,2,2,1
B. 2,2,1,1,1
C. 2,1,1,2,1
D. 2,2,1,2,1

Explanation

To balance the chemical equation \( K_2CrO_4 + HCl \rightarrow K_2Cr_2O_7 + KCl + H_2O \), we will follow these steps:

### Step 1: Write the unbalanced equation
The unbalanced equation is:
\[ K_2CrO_4 + HCl \rightarrow K_2Cr_2O_7 + KCl + H_2O \]

### Step 2: Count the number of atoms of each element on both sides
- **Left Side:**
- Potassium (K): 2
- Chromium (Cr): 1
- Oxygen (O): 4
- Hydrogen (H): 1
- Chlorine (Cl): 1

- **Right Side:**
- Potassium (K): 2 (from \( K_2Cr_2O_7 \) and \( KCl \))
- Chromium (Cr): 2
- Oxygen (O): 7 (from \( K_2Cr_2O_7 \)) + 1 (from \( H_2O \)) = 8
- Hydrogen (H): 2
- Chlorine (Cl): 1

### Step 3: Balance Chromium (Cr)
To balance chromium, we need 2 chromium atoms on the left side. We can achieve this by multiplying \( K_2CrO_4 \) by 2:
\[ 2K_2CrO_4 + HCl \rightarrow K_2Cr_2O_7 + KCl + H_2O \]

Now, recalculate:
- Left Side:
- K: 4
- Cr: 2
- O: 8
- H: 1
- Cl: 1

### Step 4: Balance Potassium (K)
Potassium is already balanced with 4 on both sides.

### Step 5: Balance Oxygen (O)
We have 8 oxygen atoms on the left side and 8 on the right side (7 from \( K_2Cr_2O_7 \) and 1 from \( H_2O \)), so oxygen is balanced.

### Step 6: Balance Hydrogen (H) and Chlorine (Cl)
We have 1 hydrogen atom on the left and 2 on the right. To balance hydrogen, we can multiply \( HCl \) by 6:
\[ 2K_2CrO_4 + 6HCl \rightarrow K_2Cr_2O_7 + 2KCl + 3H_2O \]

### Step 7: Final count of atoms
- **Left Side:**
- K: 4
- Cr: 2
- O: 8
- H: 6
- Cl: 6

- **Right Side:**
- K: 4 (from \( K_2Cr_2O_7 \) and \( 2KCl \))
- Cr: 2
- O: 8 (from \( K_2Cr_2O_7 \) and \( 3H_2O \))
- H: 6 (from \( 3H_2O \))
- Cl: 6 (from \( 2KCl \))

### Conclusion
The balanced equation is:
\[ 2K_2CrO_4 + 6HCl \rightarrow K_2Cr_2O_7 + 2KCl + 3H_2O \]

### Numerical Coefficients
The set of numerical coefficients is \( 2, 6, 1, 2, 3 \).

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