Question #a6256

When dealing with gaseous products, mroe specifically with the volume occupied by a sample of gas, you need to have some information about the pressure and temperature at which the reaction takes place.

Usually, when no mention of pressure and temperature is made, you can assume that the reactiontakes place at STP - Standard Temperature and Pressure.

At STP conditions, which imply a pressure of #"100 kPa"# and a temperature of #0^@"C"#, one mole of any ideal gas occupies exactly #"22.7 L"# - this is known as the molar volume of a gas at STP.

Basically, if you know how many moles of gas a reaction produces at STP, you can use the molar volume of a gas to determine what volume would that sample occupy.

So, the balanced chemical equation for this reaction looks like this

#"C"_3"H"_text(8(g]) + 5"O"_text(2(g]) -> color(red)(3)"CO"_text(2(g]) + 4"H"_2"O"_text((l])#

The gaseous product in this case will be carbon dioxide, #"CO"_2#.

Notice that you have a #1:color(red)(2)# mole ratio between propane and carbon dioxide, which means that the reaction will produce #color(red)(3)# times as many moles of #"CO"_2# as you have moles of #"C"_3"H"_8# that take part in the reaction.

So, if #0.1# moles of propane react completely, you will get

#0.1color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(red)(3)" moles CO"_2)/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "0.3 moles CO"_2#

This means that the volume occupied by the carbon dioxide will be

#0.3color(red)(cancel(color(black)("moles"))) * "22.7 L"/(1color(red)(cancel(color(black)("mole")))) = "6.81 L"#

Now, since you have #"1 L" = "1 dm"^3#, the answer will be

#V = color(green)("6.8 dm"^3)#

SIDE NOTE Most textbooks and onmline resources still use the old values for STP, which correspond to a pressure of 1 atm and a temperature of 0 degrees Celsius.

At these conditions, the molar volume of a gas is equal to 22.4 L. If you use that value, you will get

#0.3color(red)(cancel(color(black)("moles"))) * "22.4 L"/(1color(red)(cancel(color(black)("mole")))) = "6.72 L"#

which rounds to

#V = "6.7 dm"^3 -># option C