A chemist wants to prepare diborane by the reaction
`" " 6 LiH + 8BF_(4)to 6LiBF_(4) + B_(2)H_(6)`
If he starts with 2.0 moles each of LiH and `BF_(3)` .How many moles of `B_(2)H_(6)` can be prepared.

Answer

To solve the problem of how many moles of diborane (B₂H₆) can be prepared from the reaction given, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is:
\[ 6 \text{LiH} + 8 \text{BF}_3 \rightarrow 6 \text{LiBF}_4 + \text{B}_2\text{H}_6 \]

### Step 2: Identify the initial amounts of reactants
We are given:
- 2.0 moles of LiH
- 2.0 moles of BF₃

### Step 3: Determine the stoichiometric ratios
From the balanced equation, we see that:
- 6 moles of LiH react with 8 moles of BF₃ to produce 1 mole of B₂H₆.

### Step 4: Calculate the limiting reagent
To find the limiting reagent, we can use the stoichiometric ratios from the balanced equation.

1. For LiH:
\[ \text{Required moles of BF}_3 = \left(\frac{8 \text{ moles BF}_3}{6 \text{ moles LiH}}\right) \times 2 \text{ moles LiH = } \frac{16}{6} \text{ moles BF}_3 \approx 2.67 \text{ moles BF}_3 \]
Since we only have 2.0 moles of BF₃ available, LiH is not the limiting reagent.

2. For BF₃:
\[ \text{Required moles of LiH} = \left(\frac{6 \text{ moles LiH}}{8 \text{ moles BF}_3}\right) \times 2 \text{ moles BF}_3 = \frac{12}{8} \text{ moles LiH} = 1.5 \text{ moles LiH} \]
Since we have 2.0 moles of LiH available, BF₃ is the limiting reagent.

### Step 5: Calculate the moles of B₂H₆ produced
From the balanced equation, we know that:
- 8 moles of BF₃ produce 1 mole of B₂H₆.

Using the amount of BF₃ we have:
\[ \text{Moles of B}_2\text{H}_6 = \left(\frac{1 \text{ mole B}_2\text{H}_6}{8 \text{ moles BF}_3}\right) \times 2 \text{ moles BF}_3 = \frac{2}{8} = 0.25 \text{ moles B}_2\text{H}_6 \]

### Final Answer
The chemist can prepare **0.25 moles of B₂H₆**.

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