Balanced equation
#"C"_3"H"_8("g") + "5O"_2("g")"##stackrel("Delta)rarr##"4H"_2"O(g) + 3CO"_2("g")"#
This is a limiting reactant question. The reactant that produces the least amount of carbon dioxide is the limiting reactant and it determines the maximum amount of carbon dioxide that can be produced.
#color(red)(1.# Determine mol of each reactant by dividing by its molar mass (g/mol). Do this by multiplying by the inverse of the molar mass (mol/g).
#color(blue)(2.# Multiply mol reactant by the mol ratio between the reactant and #"CO"_2"# from the balanced equation, with #"CO"_2"# in the numerator. This will give the mol #"CO"_2"# that can be produced by the reactant.
Reactant #"O"_2"#
#color(red)(38.1color(black)cancel(color(red)("g O"_2))xx(color(red)1color(black)cancel(color(red)("mol O"_2)))/(color(red)15.999color(black)cancel(color(red)("g O"_2)))xx(color(blue)3color(black)cancel(color(blue)("mol CO"_2)))/(color(blue)5color(black)cancel(color(blue)"mol O"_2)))="1.43 mol CO"_2"#
Reactant #"C"_3"H"_8"#
#color(red)7.18color(black)cancel(color(red)("g C"_3"H"_8))xx(color(red)1color(black)cancel(color(red)("mol C"_3"H"_8)))/(color(red)44.097color(black)cancel(color(red)("g C"_3"H"_8)))xx(color(blue)3color(black)cancel(color(blue)("mol CO"_2)))/(color(blue)1color(black)cancel(color(blue)("mol C"_3"H"_8)))="0.488 mol CO"_2"#
The limiting reactant is propane. We can use the molar volume of a gas at STP to determine the volume in liters of #"CO"_2"# that can be produced.
Current STP is #0^@"C"# and #10^5# #"Pa"#. The molar volume of a gas with the current STP is #"22.711 L/mol"#.
To calculate the volume of #"0.488 mol CO"_2"# at STP, multiply mol #"CO"_2"# by the molar volume.
#"volume CO"_2=0.488color(red)cancel(color(black)("mol CO"_2))xx(22.711"L")/(color(red)cancel(color(black)("mol")))="11.1 L CO"_2"#
Many people are still using STP values of #0^@"C"# and #"1 atm"#. If that is the value you are required to use, multiply mol #"CO"_2"# by the molar volume of #"22.414 L/mol"#.
#0.488color(red)cancel(color(black)("mol CO"_2))xx(22.414"L")/(color(red)cancel(color(black)("mol")))="10.9 L CO"_2"#
