Question

When burning 180 g glucose in the presence of the 192 g of oxygen, water and carbon dioxide are produced. If 108 g of water is produced, how much carbon dioxide is produced?

Answer 1

`264` `"g CO"_2`

Explanation:

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 percent yield, 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`