20.8g of `BaCI_2` on reaction with 9.8g of `H_2SO_4` produces 7.3 g of HCI and some amount of `BaSO_4` The amount of `BaSO_4` formed is

Options

A. 23.3 g
B. 20.8 g
C. 9.8 g
D. 10.4 g

Answer

To solve the problem, we need to determine the amount of BaSO4 formed when 20.8 g of BaCl2 reacts with 9.8 g of H2SO4, producing 7.3 g of HCl. We will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction between barium chloride (BaCl2) and sulfuric acid (H2SO4) is:
\[ \text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{HCl} \]

### Step 2: Calculate the number of moles of BaCl2
To find the number of moles of BaCl2, we use the formula:
\[ \text{Number of moles} = \frac{\text{Weight (g)}}{\text{Molecular weight (g/mol)}} \]

The molecular weight of BaCl2 is calculated as follows:
- Atomic weight of Ba = 137 g/mol
- Atomic weight of Cl = 35.5 g/mol (2 Cl atoms)

So,
\[ \text{Molecular weight of BaCl}_2 = 137 + (2 \times 35.5) = 137 + 71 = 208 \text{ g/mol} \]

Now, we can calculate the number of moles of BaCl2:
\[ \text{Number of moles of BaCl}_2 = \frac{20.8 \text{ g}}{208 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Calculate the number of moles of H2SO4
Next, we calculate the number of moles of H2SO4 using its molecular weight:
- Atomic weight of H = 1 g/mol (2 H atoms)
- Atomic weight of S = 32 g/mol
- Atomic weight of O = 16 g/mol (4 O atoms)

So,
\[ \text{Molecular weight of H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \text{ g/mol} \]

Now, we can calculate the number of moles of H2SO4:
\[ \text{Number of moles of H}_2\text{SO}_4 = \frac{9.8 \text{ g}}{98 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 4: Determine the limiting reagent
From the balanced equation, we see that 1 mole of BaCl2 reacts with 1 mole of H2SO4. Since we have 0.1 moles of each, neither is in excess, and they will react completely with each other.

### Step 5: Calculate the amount of BaSO4 formed
According to the balanced equation, 1 mole of BaCl2 produces 1 mole of BaSO4. Therefore, 0.1 moles of BaCl2 will produce 0.1 moles of BaSO4.

Now, we calculate the mass of BaSO4 produced:
- Molecular weight of BaSO4:
- Atomic weight of Ba = 137 g/mol
- Atomic weight of S = 32 g/mol
- Atomic weight of O = 16 g/mol (4 O atoms)

So,
\[ \text{Molecular weight of BaSO}_4 = 137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \text{ g/mol} \]

Now, we can calculate the mass of BaSO4 produced:
\[ \text{Mass of BaSO}_4 = \text{Number of moles} \times \text{Molecular weight} = 0.1 \text{ moles} \times 233 \text{ g/mol} = 23.3 \text{ g} \]

### Final Answer
The amount of BaSO4 formed is **23.3 g**.

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