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For the following reaction, the mass of water produced from $$445\text{ g}$$ of $$\text{C}_{57}\text{H}_{110}\text{O}_6$$ is<br>$$2\text{C}_{57}\text{H}_{110}\text{O}_6(s) + 163\text{O}_2(g) \rightarrow 114\text{CO}_2(g) + 110\text{H}_2\text{O}(l)$$
Molar mass of $$\text{C}_{57}\text{H}_{110}\text{O}_6 = 57 \times 12 + 110 \times 1 + 6 \times 16 = 890\text{ g mol}^{-1}$$, so $$445\text{ g}$$ is $$0.5$$ mol. Since 2 mol give 110 mol of water, 0.5 mol give $$27.5$$ mol of water $$= 27.5 \times 18 = 495\text{ g}$$.
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