We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

**1. Gather all the information** in one place with molar masses above the formulas and everything else below them.

We start with the balanced equation.

#M_text(r):color(white)(mmm) "28,02"color(white)(mll)"32,00"color(white)(mml)"44,01"#

#color(white)(mmmmml) "2CO" +color(white)(m) "O"_2 → color(white)(m)2"CO"_2#

#"Mass/g:"color(white)(mm)25color(white)(mmll)"2,17"color(white)(mmml)36#

#"Moles:" color(white)(mmml)"0,892"color(white)(ll)"0,067 81"color(white)(mll)#

#"Divide by:"color(white)(ml)2color(white)(mmm)1#

#"Moles rxn:"color(white)(ml)"0,446"color(white)(m)"0,067 81"#

**(a) Calculate the moles of #"CO"#**

#"Moles of CO" = 25 color(red)(cancel(color(black)("g CO"))) × ("1 mol CO")/("28,02" color(red)(cancel(color(black)("g CO")))) = "1,12 mol CO"#

**(b) Calculate moles of #"O"_2#**

#"2,17"color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/("32,00" color(red)(cancel(color(black)("g O"_2)))) = "0,067 81 mol O"_2#

**2. Identify the limiting reactant**

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

#"O"_2# is the limiting reactant because it gives the fewer moles of reaction.

**3. Calculate the theoretical yield of #"CO"_2#.**

#"Theoretical yield" = "0,067 81" color(red)(cancel(color(black)("mol O"_2))) × (2 color(red)(cancel(color(black)("mol CO"_2))))/(1 color(red)(cancel(color(black)("mol O"_2)))) × "44,01 g CO"_2/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "5,97 g CO"_2#

The theoretical yield of #"CO"_2# is 5,97 g.

**4. Calculate the percent yield of #"CO"_2#**

The formula for percentage yield is

#color(blue)(bar(ul(|color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "#

#"Percent yield" = (36 color(red)(cancel(color(black)("g"))))/("5,97" color(red)(cancel(color(black)("g")))) × 100 % = 600 %#

The percent yield is 600 %.

This is an impossible result. Are you sure you gave us the correct data?