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

Explanation

To find the mass of water produced from 445 g of \( C_{57}H_{110}O_{6} \) in the given reaction, we can follow these steps:

### Step 1: Calculate the molar mass of \( C_{57}H_{110}O_{6} \)

1. **Carbon (C)**: The atomic mass is 12 g/mol. For 57 carbon atoms:
\[ 57 \times 12 = 684 \, \text{g/mol} \]

2. **Hydrogen (H)**: The atomic mass is 1 g/mol. For 110 hydrogen atoms:
\[ 110 \times 1 = 110 \, \text{g/mol} \]

3. **Oxygen (O)**: The atomic mass is 16 g/mol. For 6 oxygen atoms:
\[ 6 \times 16 = 96 \, \text{g/mol} \]

4. **Total molar mass of \( C_{57}H_{110}O_{6} \)**:
\[ 684 + 110 + 96 = 890 \, \text{g/mol} \]

### Step 2: Determine the number of moles of \( C_{57}H_{110}O_{6} \) in 445 g

Using the molar mass calculated:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{445 \, \text{g}}{890 \, \text{g/mol}} = 0.5 \, \text{moles} \]

### Step 3: Use the stoichiometry of the reaction

From the balanced equation:
\[ 2C_{57}H_{110}O_{6} + 163 O_{2} \rightarrow 114 CO_{2} + 110 H_{2}O \]
This shows that 2 moles of \( C_{57}H_{110}O_{6} \) produce 110 moles of \( H_{2}O \).

Thus, 1 mole of \( C_{57}H_{110}O_{6} \) produces:
\[ \frac{110}{2} = 55 \, \text{moles of } H_{2}O \]

### Step 4: Calculate the moles of water produced from 0.5 moles of \( C_{57}H_{110}O_{6} \)

\[ \text{Moles of } H_{2}O = 0.5 \times 55 = 27.5 \, \text{moles of } H_{2}O \]

### Step 5: Calculate the mass of water produced

The molar mass of water \( H_{2}O \) is:
\[ (2 \times 1) + (1 \times 16) = 18 \, \text{g/mol} \]

Now, calculate the mass of water produced:
\[ \text{Mass of } H_{2}O = \text{moles} \times \text{molar mass} = 27.5 \times 18 = 495 \, \text{g} \]

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

---