For the following reaction, the mass of water produced from 445 g of $$C_{57}H_{110}O_6$$ is:
$$2 \; C_{57}H_{110}O_6(s) + 163O_2(g) \rightarrow 114 \; CO_2(g) + 110H_2O(l)$$

We are given the balanced chemical equation

$$2\,C_{57}H_{110}O_6(s) \;+\; 163\,O_2(g) \;\longrightarrow\; 114\,CO_2(g) \;+\; 110\,H_2O(l)$$

and we have to find the mass of water that can be obtained from 445 g of the solid $$C_{57}H_{110}O_6$$ when there is an excess of oxygen.

First we need the molar mass of $$C_{57}H_{110}O_6$$. Using the atomic masses $$C = 12\ \text{g mol}^{-1},\; H = 1\ \text{g mol}^{-1},\; O = 16\ \text{g mol}^{-1}$$, we write

$$M(C_{57}H_{110}O_6) \;=\; 57(12) \;+\; 110(1) \;+\; 6(16).$$

Calculating term by term, we obtain

$$57(12) = 684,$$

$$110(1) = 110,$$

$$6(16) = 96.$$

Now adding these contributions,

$$M(C_{57}H_{110}O_6) = 684 + 110 + 96 = 890\ \text{g mol}^{-1}.$$

Next we convert the given mass of the compound into moles by using the formula

$$n = \dfrac{m}{M},$$

where $$n$$ is the number of moles, $$m$$ is the mass and $$M$$ is the molar mass. Substituting the numerical values,

$$n(C_{57}H_{110}O_6) = \dfrac{445\ \text{g}}{890\ \text{g mol}^{-1}} = 0.5\ \text{mol}.$$

The balanced equation shows the stoichiometric relationship between the reactant and water. We read that

$$2\ \text{mol}\ C_{57}H_{110}O_6 \;\longrightarrow\; 110\ \text{mol}\ H_2O.$$

Dividing both coefficients by 2 gives the simpler relation

$$1\ \text{mol}\ C_{57}H_{110}O_6 \;\longrightarrow\; 55\ \text{mol}\ H_2O.$$

Therefore, the number of moles of water produced from the 0.5 mol of the compound is

$$n(H_2O) = 0.5\ \text{mol}\times 55 = 27.5\ \text{mol}.$$

Finally, we convert the moles of water into mass by using the molar mass of water, which is $$18\ \text{g mol}^{-1}$$. Thus,

$$m(H_2O) = n(H_2O)\times M(H_2O) = 27.5\ \text{mol}\times 18\ \text{g mol}^{-1}.$$

Multiplying, we find

$$m(H_2O) = 495\ \text{g}.$$

Hence, the correct answer is Option D.