1. Given reaction

$2C_{57}H_{110}O_6 + 163O_2 \rightarrow 114CO_2 + 110H_2O$

We need the mass of water formed from 445 g of $C_{57}H_{110}O_6$.

2. Find molar mass of $C_{57}H_{110}O_6$

Using $C=12$, $H=1$, $O=16$:

$M = 57(12) + 110(1) + 6(16)$

$= 684 + 110 + 96 = 890 \text{ g mol}^{-1}$

So,

$1 \text{ mol of } C_{57}H_{110}O_6 = 890 \text{ g}$

Therefore,

$445 \text{ g} = \frac{445}{890} = 0.5 \text{ mol}$

3. Use stoichiometry from balanced equation

From the reaction:

$2 \text{ mol } C_{57}H_{110}O_6 \rightarrow 110 \text{ mol } H_2O$

So,

$1 \text{ mol } C_{57}H_{110}O_6 \rightarrow 55 \text{ mol } H_2O$

Hence,

$0.5 \text{ mol } C_{57}H_{110}O_6 \rightarrow 0.5 \times 55 = 27.5 \text{ mol } H_2O$

4. Convert moles of water to mass

Molar mass of water:

$M(H_2O)=18 \text{ g mol}^{-1}$

Mass of water formed:

$27.5 \times 18 = 495 \text{ g}$

5. Check options

• A: $490$ g
• B: $445$ g
• C: $495$ g
• D: $890$ g

So the correct option is:

$\boxed{\text{C: }495\text{ g}}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They match.