Question

Question #9f7a6

Answer 1

Your reaction will produce `"0.270"` moles of carbon dioxide.

First, start with the balanced chemical equation for the combustion of propane

`C_3H_(8(g)) + 5O_(2(g)) -> 3CO_(2(g)) + 4H_2O_((l))`

Notice the `"1:3"` mole ratio that exists between propane and carbon dioxide - for every 1 mole of propane that reacts, 3 moles of carbon dioxide are produced.

In other words, regardless of how many moles of propane react, you'll always have 3 times more moles of carbond ioxide produced.

Now, in order to determine how many moles of propane rected, you must use the ideal gas law equation, `PV = nRT`.

`PV = nRT => n = (PV)/(RT)`

`n_("propane") = (69.6/101.325"atm" * 3".18 L")/(0.082("atm" * "L")/("mol" * "K") * (273.15 + 24)"K") = "0.0900 moles"`

SIDE NOTE If you want to use the value of R = 0.082, expressed in atm L/mol K, do not forget to transform the pressure from kPa to atm and the temperature from Celsius to Kelvin.

Since you know how many moles of propane reacted, you automatically know - the mole ratio tells you - how many moles of carbond sioxide were produced

`"0.0900 moles propane" * "3 moles carbon dioxide"/"1 mole propane" = "0.270 moles carbon dioxide"`