Given the equation #C_2H_6(g) + O_2(g) -> CO_2 (g) + H_2O(g)# (not balanced), what is the number of liters of #CO_2# formed at STP when 240.0 grams of #C_2H_6# is burned in excess oxygen gas?
1 Answer
Explanation:
The first thing to do here is make sure that you have a balanced chemical equation. The equation given to you is actually unbalanced, so focus on writing a balanced version
#color(red)(2)"C"_2"H"_text(6(g]) + 7"O"_text(2(g]) -> color(purple)(4)"CO"_text(2(g]) + 6"H"_2"O"_text((g])#
Now, the problem provides you with a mass of ethane,
As you know, STP conditions are defined as a temperature of
So, what this means is that if you know how many moles of carbon dioxide are produced by the reaction, you can use the molar volume of the gas to find the requested volume.
Use the molar mass of ethane to determine how many moles you'd get in that
#240.0 color(red)(cancel(color(black)("g"))) * ("1 mole C"_2"H"_6)/(30.07color(red)(cancel(color(black)("g")))) = "7.9814 moles C"_2"H"_6#
Now, notice that you have a
This means that the reaction will produce
#7.9814 color(red)(cancel(color(black)("moles C"_2"H"_6))) * (color(purple)(4)" moles CO"_2)/(color(red)(2)color(red)(cancel(color(black)("moles C"_2"H"_6)))) = "15.9628 moles CO"_2#
So, if one mole of any ideal gas occupies
#15.9628color(red)(cancel(color(black)("moles CO"_2))) * "22.7 L"/(1color(red)(cancel(color(black)("mole CO"_2)))) = color(green)("362.4 L")#
The answer is rounded to four sig figs, the number of sig figs you have for the mass of ethane.