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We're asked to find the mass of $"CO"_2$ produced in a given reaction, given that $108$ $"g H"_2"O"$ is formed.

What we can do first is write the balanced chemical equation for this reaction:

$"C"_6"H"_12"O"_6(s) + 6"O"_2(g) rarr 6"CO"_2(g) + 6"H"_2"O"(g)$

We really only needed the amount of water produced, because that would already factor in the limiting reactant and , so we'll use $108$ $"g H"_2"O"$ in our calculations.

Convert mass of water to moles using its molar mass:

$108cancel("g H"_2"O")((1color(white)(l)"mol H"_2"O")/(18.015cancel("g H"_2"O"))) = color(red)(5.99$ $color(red)("mol H"_2"O"$

Now, using the coefficients of the chemical equation, we can figure out the relative number of moles of $"CO"_2$ that form:

$color(red)(5.99)cancel(color(red)("mol H"_2"O"))((6color(white)(l)"mol CO"_2)/(6cancel("mol H"_2"O"))) = 5.99color(white)(l)"mol CO"_2$

Finally, we use the molar mass of carbon dioxide to find the number of grams:

$5.99cancel("mol CO"_2)((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(ul(264color(white)(l)"g CO"_2$

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