What is the mass of iodine required to react with 100g of `Ba(SCN)_(2)` according to the following reaction
`Ba(SCN)_(2) + 6I_(2) rarr CN^(-) + SO_(4)^(2-) + I^(-)`

Options

A. 150g
B. 600g
C. 400g
D. 100g

Answer

To find the mass of iodine required to react with 100g of \( Ba(SCN)_2 \) according to the reaction:

\[ Ba(SCN)_2 + 6I_2 \rightarrow CN^- + SO_4^{2-} + I^- \]

we will follow these steps:

### Step 1: Determine the molar mass of \( Ba(SCN)_2 \)

The molar mass of \( Ba(SCN)_2 \) can be calculated by adding the atomic masses of its constituent elements:

- Barium (Ba): 137.33 g/mol
- Sulfur (S): 32.07 g/mol
- Carbon (C): 12.01 g/mol
- Nitrogen (N): 14.01 g/mol

The formula \( Ba(SCN)_2 \) contains:
- 1 Ba
- 2 S
- 2 C
- 2 N

Calculating the molar mass:
\[ \text{Molar mass of } Ba(SCN)_2 = 137.33 + 2(32.07) + 2(12.01) + 2(14.01) = 137.33 + 64.14 + 24.02 + 28.02 = 253.51 \, \text{g/mol} \]

### Step 2: Calculate the number of moles of \( Ba(SCN)_2 \)

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given mass of \( Ba(SCN)_2 \) = 100 g, we can calculate the number of moles:
\[ \text{Number of moles of } Ba(SCN)_2 = \frac{100 \, \text{g}}{253.51 \, \text{g/mol}} \approx 0.3945 \, \text{moles} \]

### Step 3: Determine the number of moles of \( I_2 \) required

From the balanced equation, 1 mole of \( Ba(SCN)_2 \) reacts with 6 moles of \( I_2 \). Therefore, for \( 0.3945 \) moles of \( Ba(SCN)_2 \):
\[ \text{Moles of } I_2 = 0.3945 \, \text{moles} \times 6 = 2.367 \, \text{moles} \]

### Step 4: Calculate the mass of \( I_2 \)

The molar mass of \( I_2 \) (iodine) is:
\[ \text{Molar mass of } I_2 = 2 \times 126.90 \, \text{g/mol} = 253.80 \, \text{g/mol} \]

Now, we can find the mass of \( I_2 \):
\[ \text{Mass of } I_2 = \text{Number of moles} \times \text{Molar mass} = 2.367 \, \text{moles} \times 253.80 \, \text{g/mol} \approx 600.77 \, \text{g} \]

### Final Answer

The mass of iodine required to react with 100g of \( Ba(SCN)_2 \) is approximately **600 grams**.

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