Question

Question #f19e7

Answer 1

I assume you're referring to the combustion of glucose, `C_6H_12O_6`.

Start with the balanced chemical equation for this reaction

`C_6H_12O_6 + 6O_2 -> color(red)(6)CO_2 + 6H_2O`

You know that you're reaction produced 20 L (`"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 "` `CO_2`

Now take a look at the balanced chemical equation again. Notice that you have `1:color(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 `CO_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.