# Question f19e7

Apr 26, 2015

I assume you're referring to the combustion of glucose, ${C}_{6} {H}_{12} {O}_{6}$.

${C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2} \to \textcolor{red}{6} C {O}_{2} + 6 {H}_{2} O$

You know that you're reaction produced 20 L ($\text{1 dm"^3 = "1 L}$, so I'll just use liters from now on) of carbon dioxide at STP.

STP conditions imply a pressure of 100 kPa and a temperature of 273.15 K. Under these conditions, 1 mole of any ideal gas occupies exactly 22.7 L - this is know as the molar volume of a gas at STP.

Since you've collected less than 22.7 L, you can expect to have less than 1 mole of carbon dioxide produced.

20cancel("L") * "1 mole"/(22.7cancel("L")) = "0.881 moles " $C {O}_{2}$

Now take a look at the balanced chemical equation again. Notice that you have $1 : \textcolor{red}{6}$ mole ratio between glucose and carbon dioxide. What that tells you is that, regardless of how many moles of glucose react, you'll always produce 6 times more moles of carbon dioxide.

Since you know how many moles of carbon dioxide were produced, you can determine the number of moles of glucose that reacted by

0.881cancel("moles "CO_2) * "1 mole glucose"/(color(red)(6) cancel("moles " CO_2)) = "0.1468 moles glucose"

Now just use glucose's molar mas to see how many grams would contain that many moles

0.1468cancel("moles glucose") * "180.16 g"/(1cancel("mole glucose")) = "26.4 g"

If you round this off to one sig fig, the number of sig figs given for the volume of $C {O}_{2}$ produced, the answer will be

m_"glucose" = color(green)("30 g")#

SIDE NOTE More often than not, the molar volume of a gas at STP is given as 22.4 L. That value is based on the old definition of STP, which implied a pressure of 1 atm and a temperature of 273.15 K.

The current value for the molar volume of a gas is 22.7 L. If your teacher or instructor wanted you to use the old value, just redo the calculations with 22.4 L.