# 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?

Jul 29, 2017

$264$ ${\text{g CO}}_{2}$

#### Explanation:

We're asked to find the mass of ${\text{CO}}_{2}$ produced in a given reaction, given that $108$ $\text{g H"_2"O}$ is formed.

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

$\text{C"_6"H"_12"O"_6(s) + 6"O"_2(g) rarr 6"CO"_2(g) + 6"H"_2"O} \left(g\right)$

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 $108$ $\text{g H"_2"O}$ 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 ${\text{CO}}_{2}$ 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