Step 1 - Reading the chemical reaction
The given chemical reaction is:
[tex]2C_3H_7OH+9O_2\rightarrow6CO_2+8H_2O[/tex]We can read this reaction as:
2 moles of C3H7OH react with 9 moles of O2 to produce 6 moles of CO2 and 8 moles of H2O
This is what we call theoretical reaction, i.e., a theoretical proportion that will be respected whenever this reaction happens.
As the exercise is specifically asking about the reactants, we can further simplify this statement to:
2 moles of C3H7OH react with 9 moles of O2
Step 2 - Converting this proportion to grams
We can convert moles to grams by multiplying the number of moles by the molar mass of each substance (32 g/mol for O2; 60 g/mol for C3H7OH):
[tex]\begin{gathered} C_3H_7OH\rightarrow2moles\times60g/mol=120g \\ \\ O_2\rightarrow9moles\times32g/mol=288g \end{gathered}[/tex]Therefore, we obtain the following relation:
288g of O2 react with 120g of C3H7OH
This is a fixed proportion. Whenever this reaction happens, it will respect this proportion. We can use it thus to find the limiting reactant.
Step 3 - Finding the limiting reactant
To discover the limiting reactant, we first must discover how much O2 is needed to completely react with 8 grams of C3H7OH:
[tex]\begin{gathered} 288g\text{ of O2 react with ----- 120g of C3H7OH} \\ xg\text{ of O2 would react with ---- 8g of C3H7OH} \\ \\ x=\frac{288\times8}{120}=19.2g\text{ of O2} \end{gathered}[/tex]Therefore, in order to completely consume 8 grams of C3H7OH, we would need 19.2g of O2. Since there less O2 than needed, it's the limiting reactant.
Answer: C3H7OH is in excess. Therefore, O2 is the limiting reactant.