#"2.59 g"# of sucrose will produce #"4.00 g"# of carbon dioxide when reacted completely. I assume you can use the molar masses of sucrose and carbon dioxide, since you can easily determine them just by looking at the periodic table.
So, the combustion of sucrose has this balanced chemical equation
#C_12H_22O_11 + 12O_2 -> 11H_2O + 12CO_2#
Since no information about the amount of oxygen is given, you can safely assume that it will not act as a limiting reagent, which implies that you either have just enough of it, or you have it in excess.
Use the balanced equation to determine how much carbon dioxide will be produced in the reaction. This is done by looking at the mole ratio between sucrose and carbon dioxide.
Notice that 1 mole of sucrose will produce 12 moles of #CO_2#, which translates into a #"1:12"# mole ratio between the two. However many moles of sucrose you'll use in the reaction, 12 times more moles #CO_2# will be produced.
The number of sucrose moles that react is
#"2.59 g" * ("1 mole")/("342.3 g") = "0.00757 moles sucrose"#
This means that you'll produce
#"0.00757 moles sucrose" * ("12 moles "CO_2)/("1 mole sucrose") = "0.0908 moles "CO_2#
Now use carbon dioxide's molar mass to go from moles to grams
#"0.0908 moles" * ("44.0 g")/("1 mole") = "4.00 g"#
If you're not allowed to use the actual molar masses, just go generic. If #M_("sucrose")# is the molar mass of sucrose and #M_(CO_2)# is the molar mass of carbon dioxide, you'll have
#"2.59 g" * ("1 mole")/(M_("sucrose")) = "2.59 g"/M_("sucrose")# #-># moles of sucrose
#"2.59 g"/M_("sucrose") * ("12 moles "CO_2)/("1 mole sucrose") = "31.1 g"/(M_("sucrose"))# #-># moles of #CO_2#
#"31.1 g"/M_("sucrose") * (M_(CO_2))/("1 mole") = "31.1 g" * M_(CO_2)/(M_("sucrose")# #-># grams of #CO_2#
If you use the actual molar masses, the result will be identical #"4.00 g"#.
