For the following reaction, the mass of water produced from 445 g of `C_(57)H_(110)O_(6)` is :
`2C_(57)H_(110)O_(6)(s)+163O_(2)(g) to 114CO_(2)(g)+110H_(2)O(l)`

Options

A. 445 g
B. 495 g
C. 890 g
D. 490 g

Explanation

To solve the problem of finding the mass of water produced from the reaction of 445 g of \( C_{57}H_{110}O_{6} \), we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation is given as:
\[ 2C_{57}H_{110}O_{6}(s) + 163O_{2}(g) \rightarrow 114CO_{2}(g) + 110H_{2}O(l) \]

### Step 2: Calculate the molar mass of \( C_{57}H_{110}O_{6} \)
To find the molar mass, we sum the atomic masses of each element in the compound:
- Carbon (C): 57 atoms × 12.01 g/mol = 684.57 g/mol
- Hydrogen (H): 110 atoms × 1.008 g/mol = 110.88 g/mol
- Oxygen (O): 6 atoms × 16.00 g/mol = 96.00 g/mol

Calculating the total:
\[ \text{Molar mass of } C_{57}H_{110}O_{6} = 684.57 + 110.88 + 96.00 = 891.45 \text{ g/mol} \]

### Step 3: Calculate the number of moles of \( C_{57}H_{110}O_{6} \)
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{445 \text{ g}}{891.45 \text{ g/mol}} \approx 0.5 \text{ moles} \]

### Step 4: Determine the moles of water produced
From the balanced equation, we see that 2 moles of \( C_{57}H_{110}O_{6} \) produce 110 moles of \( H_{2}O \). Therefore, 1 mole of \( C_{57}H_{110}O_{6} \) produces:
\[ \frac{110 \text{ moles } H_{2}O}{2 \text{ moles } C_{57}H_{110}O_{6}} = 55 \text{ moles } H_{2}O \]
Thus, for 0.5 moles of \( C_{57}H_{110}O_{6} \):
\[ \text{Moles of } H_{2}O = 0.5 \text{ moles } C_{57}H_{110}O_{6} \times 55 \text{ moles } H_{2}O = 27.5 \text{ moles } H_{2}O \]

### Step 5: Calculate the mass of water produced
Using the molar mass of water (\( H_{2}O \)), which is approximately 18 g/mol:
\[ \text{Mass of } H_{2}O = \text{moles} \times \text{molar mass} = 27.5 \text{ moles} \times 18 \text{ g/mol} = 495 \text{ g} \]

### Final Answer
The mass of water produced from 445 g of \( C_{57}H_{110}O_{6} \) is **495 g**.

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