# The combustion of glucose "C"_6"H"_12"O"_6 produces carbon dioxide and water. What mass of glucose will produce 44 g of carbon dioxide?

Sep 5, 2017

The mass of glucose required is 30 g.

${M}_{\textrm{r}} : \textcolor{w h i t e}{m l} 180.16 \textcolor{w h i t e}{m m m m m m m m m m l} 44.01$
$\textcolor{w h i t e}{m m m} \text{C"_6"H"_12"O"_6 + "6O"_2 → "6CO"_2 + 6"H"_2"O}$

Step 2. Calculate the moles of $\text{CO"_2}$

$\text{Moles of CO"_2 = 44 color(red)(cancel(color(black)("g CO"_2))) × ("1 mol CO"_2)/(44.01color(red)(cancel(color(black)("g CO"_2)))) = "1.00 mol CO"_2}$

Step 3. Calculate the moles of ${\text{C"_6"H"_12"O}}_{6}$

${\text{Moles of C"_6"H"_12"O"_6 = 1.00 color(red)(cancel(color(black)("mol CO"_2))) × (1 "mol C"_6"H"_12"O"_6)/(6color(red)(cancel(color(black)("mol CO"_2)))) = "0.167 mol C"_6"H"_12"O}}_{6}$

Step 4. Calculate the mass of ${\text{C"_6"H"_12"O}}_{6}$

${\text{Mass of C"_6"H"_12"O"_6 = 0.167 color(red)(cancel(color(black)("mol C"_6"H"_12"O"_6))) × ("180.16 g C"_6"H"_12"O"_6)/(1 color(red)(cancel(color(black)("mol C"_6"H"_12"O"_6)))) = "30 g C"_6"H"_12"O}}_{6}$