`H_(2)underline(S_(2))O_(7)+H_(2)O to H_(2)SO_(4)`

Options

A. If product is oxy acid with -ic suffix.
B. If product is oxy acid with -ous suffix
C. If product are two oxy acids one with -ic suffix and otherone with -ous suffix.
D. If product is not oxy acid, neither with -ic suffix nor with -ous suffix

Answer

To solve the reaction \( H_2S_2O_7 + H_2O \rightarrow H_2SO_4 \), we will analyze the reactants and the products step by step.

### Step 1: Identify the Reactants
The reactants in the equation are:
- \( H_2S_2O_7 \) (disulfuric acid or oleum)
- \( H_2O \) (water)

### Step 2: Understand the Reaction
When \( H_2S_2O_7 \) reacts with water, it undergoes hydrolysis. Hydrolysis is a reaction with water that results in the formation of an acid. In this case, \( H_2S_2O_7 \) reacts with water to produce sulfuric acid.

### Step 3: Write the Product
The product of the reaction is \( H_2SO_4 \) (sulfuric acid). This is formed by the addition of water to \( H_2S_2O_7 \).

### Step 4: Confirm the Type of Reaction
This reaction can be classified as a hydrolysis reaction, where a compound reacts with water to form an acid. In this case, the compound \( H_2S_2O_7 \) is an oxyacid that produces another oxyacid, \( H_2SO_4 \).

### Step 5: Identify the Oxyacid
An oxyacid is defined as an acid that contains oxygen, hydrogen, and another element. In this case:
- \( H_2SO_4 \) contains sulfur (S), oxygen (O), and hydrogen (H).
- The anion formed from \( H_2SO_4 \) is \( SO_4^{2-} \), which is called sulfate.

### Step 6: Naming the Oxyacid
Since the anion \( SO_4^{2-} \) ends with "ate", the corresponding acid is named using the suffix "ic". Therefore, \( H_2SO_4 \) is named sulfuric acid.

### Conclusion
The reaction \( H_2S_2O_7 + H_2O \rightarrow H_2SO_4 \) results in the formation of sulfuric acid, which is an oxyacid with the suffix "ic".

### Final Answer
The correct option is:
- If the product is an oxyacid with "ic" suffix, which corresponds to sulfuric acid.

---