0.7 moles of potassium sulphate is allowed to react with 0.9 moles of barium chloride in aqueous solutions. The number of moles of the substance precipitated in the reactions is

Options

A. 1.4 moles of potassium chloride
B. 0.7 moles of barium sulphate
C. 1.6 moles of potassium chloride
D. 1.6 moles of barium sulphate

Explanation

To solve the problem of how many moles of the substance precipitated when 0.7 moles of potassium sulfate reacts with 0.9 moles of barium chloride, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The reaction between potassium sulfate (K₂SO₄) and barium chloride (BaCl₂) can be written as:
\[ K_2SO_4 (aq) + BaCl_2 (aq) \rightarrow BaSO_4 (s) + 2 KCl (aq) \]
This equation shows that 1 mole of K₂SO₄ reacts with 1 mole of BaCl₂ to produce 1 mole of barium sulfate (BaSO₄) as a precipitate.

2. **Identify the Limiting Reagent:**
- From the balanced equation, we see that 1 mole of K₂SO₄ reacts with 1 mole of BaCl₂.
- We have 0.7 moles of K₂SO₄ and 0.9 moles of BaCl₂.
- Since K₂SO₄ is present in a smaller amount (0.7 moles), it will be the limiting reagent.

3. **Calculate the Moles of Precipitate Formed:**
- According to the balanced equation, 1 mole of K₂SO₄ produces 1 mole of BaSO₄.
- Therefore, 0.7 moles of K₂SO₄ will produce 0.7 moles of BaSO₄.
- Thus, the number of moles of the precipitate (BaSO₄) formed is 0.7 moles.

4. **Conclusion:**
The number of moles of the substance precipitated (BaSO₄) in the reaction is **0.7 moles**.

### Summary of the Solution:
- The balanced equation shows the stoichiometry of the reaction.
- K₂SO₄ is the limiting reagent.
- The amount of BaSO₄ precipitated is directly proportional to the amount of K₂SO₄ used.