There are 100.0 grams of each reactant available determine the limiting reactant in this equation

Since you have not included the chemical reaction I will explain you in detail.

1) To determine the limiting agent you need two things:

- the balanced chemical equation

- the amount of every reactant involved as per the chemical equation

2) The work is:

- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.

- determine the number of moles of each reactant with this formula:

number of moles = (mass in grams) / (molar mass)

- set the proportion with the two ratios (theoretical moles and actual moles)

- compare which reactant is below than the stated by the theoretical ratio.

3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:

i) Chemical equation: H₂ + O₂ → H₂O

ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O

iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂

iv) Covert 100 g of H₂ into number of moles

n = 100g / 2g/mol = 50 mol of H₂

v) Convert 100 g of O₂ to moles: 

n = 100 g / 32 g/mol = 3.125 mol

vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂

vii) Compare the two ratios:

2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂

Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.

Therefore, the O₂ is the limiting reactant in this example.

Because it was a dramatically big guy

What do you need with it’s not a question your just stating what it says