We're asked to find the mass of #"CO"_2# that forms when #30# #"g CO"# are burned in excess oxygen.
The equation for this reaction is
#2"CO" (g) + "O"_2(g) rarr 2"CO"_2(g)#
Let's first calculate the number of moles of carbon monoxide, using its molar mass, #28.01# #"g/mol"#:
#30cancel("g CO")((1color(white)(l)"mol CO")/(28.01cancel("g CO"))) = color(red)(1.07# #color(red)("mol CO"#
Since #"CO"# and #"CO"_2# have the same coefficients, the relative number of moles will be the same:
#color(red)(1.07)color(white)(l)color(red)("mol CO") = color(green)(1.07)color(white)(l)color(green)("mol CO"_2#
Lastly, let's use the molar mass of #"CO"_2# (#44.01# #"g/mol"#) to calculate the theoretical mass yield of carbon dioxide:
#color(green)(1.07)cancel(color(green)("mol CO"_2))((44.01color(white)(l)"g CO"_2)/(1cancel("mol CO"_2))) = color(blue)(47# #color(blue)("g CO"_2#
rounded to #2# significant figures.
Thus, for this reaction, if #30# grams of carbon monoxide are burned in air, #color(blue)(47# #sfcolor(blue)("grams of CO"_2# can form (this is the theoretical yield, the maximum amount that can form).
