# 1)how many grams of C6H5CO2H are produced from 8.00 x 10^25 atoms of oxygen? 2)If 2.6g of C6H5CH3 and 3.2g of O2 are allowed to react which will be limiting? 3)how many moles of C6H5CO2H will be formed(theoretical yield)?

Feb 13, 2018

1) 10,815.06g
2)C6H5CH3 Limiting
3)0.0282184mol C6H5CO2H

#### Explanation:

1) By utilizing Avagadro's number ($6.02 \cdot {10}^{23}$) atoms/mol) we can determine the number of moles of oxygen we have.

$8.00 \cdot {10}^{25}$ Atoms/$6.02 \cdot {10}^{23}$ atoms/mol = 132.84moles

If we look at the chemical formula of C6H5CO2H we can determine that 3 moles of oxygen are required for every 2 moles of this compound, therefore; 132.84moles ${O}_{2}$ * $2 {C}_{6} {H}_{5} C {O}_{2} H$ / $3 {O}_{2}$ = $88.56$Mols of ${C}_{6} {H}_{5} C {O}_{2} H$

By multiplying by the molar mass of the product (In g/mol) we get the amount of product in grams.

2) When we look at the balanced reaction:

$2 {C}_{6} {H}_{5} C {H}_{3} + 3 {O}_{2} \to 2 {C}_{6} {H}_{5} C {O}_{2} H + 2 {H}_{2} O$

We see that for every 2 moles of $2 {C}_{6} {H}_{5} C {O}_{2} H$ we require 2 moles of ${C}_{6} {H}_{5} C {H}_{3}$

or

3 moles of ${O}_{2}$ (6moles of O) for every 2 moles of product.

By taking this into account we can determine which will be the limiting reagent based on the amount of product produced!

(2.6g ${C}_{6} {H}_{5} C {H}_{3}$ / (92.1384g/mol)) * (2 mol $2 {C}_{6} {H}_{5} C {O}_{2} H$ / 2mol ${C}_{6} {H}_{5} C {H}_{3}$) equals 0.028218 moles of product.

This can be repeated with ${O}_{2}$ (approx 32 g/mol)

(3.2g ${O}_{2}$ / (32g/mol)) * (2 mol ${C}_{6} {H}_{5} C {O}_{2} H$ / $3 m o l {O}_{2}$ ) which equals 0.06666

3)

To answer this question we must first understand what a limiting reagent is;

"The limiting reagent (or limiting reactant, LR) in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it." - Wikipedia

Therefore by looking at the above quantities, we see that ${C}_{6} {H}_{5} C {H}_{3}$ is the limiting reagent due to this reagent being consumed to completion first.

This results in a total amount of moles of our product being produced as 0.028218 moles