Consider the following sequence of reactions:
`B_(2)O_(3)+CaF_(2)+H_(2)SO_(4) to (A) + (B) + H_(2)O`
`(A)+ LiAlH_(4) to ( C) (g) +(D) + LiF`

Explanation

To solve the given sequence of reactions, we will break down the steps systematically.

### Step 1: Identify the first reaction
The first reaction is given as:
\[ B_2O_3 + CaF_2 + H_2SO_4 \rightarrow (A) + (B) + H_2O \]

**Analysis:**
- When boron trioxide (\(B_2O_3\)) reacts with calcium fluoride (\(CaF_2\)) and sulfuric acid (\(H_2SO_4\)), it typically forms boron trifluoride (\(BF_3\)) and calcium sulfate (\(CaSO_4\)) along with water.

**Balanced Equation:**
To balance the equation, we can write:
\[ B_2O_3 + 3CaF_2 + 3H_2SO_4 \rightarrow 2BF_3 + 3CaSO_4 + 3H_2O \]

From this reaction:
- Compound \(A\) is \(BF_3\)
- Compound \(B\) is \(CaSO_4\)

### Step 2: Identify the second reaction
The second reaction is:
\[ (A) + LiAlH_4 \rightarrow (C) + (D) + LiF \]

**Analysis:**
- Here, \(A\) is \(BF_3\), and it reacts with lithium aluminum hydride (\(LiAlH_4\)). The reaction typically produces diborane (\(B_2H_6\)), aluminum trifluoride (\(AlF_3\)), and lithium fluoride (\(LiF\)).

**Balanced Equation:**
To balance this equation, we can write:
\[ 4BF_3 + 3LiAlH_4 \rightarrow 2B_2H_6 + 3AlF_3 + 3LiF \]

From this reaction:
- Compound \(C\) is \(B_2H_6\) (gaseous)
- Compound \(D\) is \(AlF_3\)

### Summary of Compounds
- \(A = BF_3\)
- \(B = CaSO_4\)
- \(C = B_2H_6\) (gaseous)
- \(D = AlF_3\)

### Final Answer
- Compound \(A\) is \(BF_3\)
- Compound \(B\) is \(CaSO_4\)
- Compound \(C\) is \(B_2H_6\)
- Compound \(D\) is \(AlF_3\)

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