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 100 g 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: Calculate the molar mass of \( Ba(SCN)_2 \)

The molar mass of \( Ba(SCN)_2 \) can be calculated as follows:
- 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 for \( Ba(SCN)_2 \) consists of:
- 1 Ba
- 2 S
- 2 C
- 6 N

Calculating the molar mass:
\[ \text{Molar mass of } Ba(SCN)_2 = 137.33 + 2(32.07) + 2(12.01) + 6(14.01) = 137.33 + 64.14 + 24.02 + 84.06 = 309.55 \, \text{g/mol} \]

### Step 2: Determine the number of moles of \( Ba(SCN)_2 \) in 100 g

Using the molar mass calculated:
\[ \text{Number of moles of } Ba(SCN)_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \, \text{g}}{309.55 \, \text{g/mol}} \approx 0.323 \, \text{mol} \]

### Step 3: Use the stoichiometry of the reaction

From the balanced equation, 1 mole of \( Ba(SCN)_2 \) reacts with 6 moles of \( I_2 \). Therefore, the number of moles of \( I_2 \) required for 0.323 moles of \( Ba(SCN)_2 \) is:
\[ \text{Moles of } I_2 = 0.323 \, \text{mol} \times 6 = 1.938 \, \text{mol} \]

### 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 calculate the mass of \( I_2 \):
\[ \text{Mass of } I_2 = \text{moles} \times \text{molar mass} = 1.938 \, \text{mol} \times 253.80 \, \text{g/mol} \approx 491.11 \, \text{g} \]

### Final Answer

The mass of iodine required to react with 100 g of \( Ba(SCN)_2 \) is approximately **491.11 g**.

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