The answer is #"4.24 L CO"_2"#.
Start with a balanced equation:
#"2 C"_4"H"_10" + 13 O"_2"# #rarr# #"8 CO"_2" + 10 H"_2"O"#
The molar mass of butane, #"C"_4"H"_10"# is #"58.12 g/mol"#. http://en.wikipedia.org/wiki/Butane
Determine the number of moles in #"2.75 g C"_4"H"_10"#.
#"2.75 g C"_4"H"_10"# x #"1 mol"/"58.12 g"# = #"0.0473 mol C"_4"H"_10"#
Multiplly the calculated number of moles of butane times the mole ratio of butane to carbon dioxide, such that carbon dioxide is on top.
#"0.0473 mol butane"# x #"8 mol carbon dioxide"/"2 mol butane"# = #"0.189 mol CO"_2"#
The molar volume of a gas at STP is #"22.414 L/mol"#. Multiply the calculated mol of carbon dioxide times #"22.414 L"/"1 mol"#.
#"0.189 mol CO"_2"# x #"22.414 L"/"1 mol"# = #"4.24 L CO"_2"#
NOTE: The actual molar volume at combustion temperatures is not going to be the same as the molar volume at STP, which is #0^("o")"C"# and #"1 atm"#, so this is really more of an approximation.
