For the following reaction, the moles of required to produce 990 gm `H_2O` are __________.
`2C_(57)H_(110)O_(6) + 163O_(2)(g) to 114CO_(2)(g) + 110 H_(2)O(l)`

Explanation

To determine the number of moles of \( C_{57}H_{110}O_{6} \) required to produce 990 grams of \( H_2O \), we can follow these steps:

### Step 1: Understand the Reaction
The balanced chemical equation is:
\[ 2 C_{57}H_{110}O_{6} + 163 O_{2} \rightarrow 114 CO_{2} + 110 H_{2}O \]

From the equation, we can see that 2 moles of \( C_{57}H_{110}O_{6} \) produce 110 moles of \( H_{2}O \).

### Step 2: Calculate Moles of Water
First, we need to calculate the number of moles of \( H_{2}O \) in 990 grams. The molar mass of \( H_{2}O \) can be calculated as follows:
- Molar mass of \( H_{2}O \) = \( 2 \times 1 + 16 = 18 \, \text{g/mol} \)

Now, we can calculate the number of moles of \( H_{2}O \):
\[ \text{Moles of } H_{2}O = \frac{\text{mass}}{\text{molar mass}} = \frac{990 \, \text{g}}{18 \, \text{g/mol}} = 55 \, \text{moles} \]

### Step 3: Relate Moles of Water to Moles of \( C_{57}H_{110}O_{6} \)
From the balanced equation, we know that:
- 110 moles of \( H_{2}O \) are produced from 2 moles of \( C_{57}H_{110}O_{6} \).

To find out how many moles of \( C_{57}H_{110}O_{6} \) are needed to produce 55 moles of \( H_{2}O \), we can set up a proportion:
\[ \frac{2 \, \text{moles of } C_{57}H_{110}O_{6}}{110 \, \text{moles of } H_{2}O} = \frac{x \, \text{moles of } C_{57}H_{110}O_{6}}{55 \, \text{moles of } H_{2}O} \]

Cross-multiplying gives:
\[ 110x = 2 \times 55 \]
\[ 110x = 110 \]
\[ x = 1 \]

### Conclusion
Thus, the number of moles of \( C_{57}H_{110}O_{6} \) required to produce 990 grams of \( H_{2}O \) is **1 mole**.

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