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)?

Question

help me please

Noah Aldrich · Accepted Answer

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

1) By utilizing Avagadro's number (#6.02*10^23#) atoms/mol) we can determine the number of moles of oxygen we have. #8.00*10^25# Atoms/#6.02*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# * #2C_6H_5CO_2H# / #3O_2# = #88.56#Mols of #C_6H_5CO_2H # We may calculate the amount of product in grams by multiplying by the product's molar mass (in g/mol). 2) Upon examining the equilibrium response: #2C_6H_5CH_3 + 3O_2 -> 2C_6H_5CO_2H + 2H_2O# We see that for every 2 moles of #2C_6H_5CO_2H# we require 2 moles of #C_6H_5CH_3# or 3 moles of #O_2# (6moles of O) for every 2 moles of product. Considering this, we can ascertain the limiting reagent by considering the quantity of product generated! (2.6g #C_6H_5CH_3# / (92.1384g/mol)) * (2 mol #2C_6H_5CO_2H# / 2mol #C_6H_5CH_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_6H_5CO_2H# / #3 mol O_2# ) which equals 0.06666 3) Prior to responding, let's clarify 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." - According to Wikipedia, Therefore by looking at the above quantities, we see that #C_6H_5CH_3# is the limiting reagent due to this reagent being consumed to completion first. As a consequence, 0.028218 moles of our product were produced in total.

Jose Ayala · Answer

undefined 1) Approximately 41.5 grams of C6H5CO2H are produced from 8.00 x 10^25 atoms of oxygen.
2) C6H5CH3 will be the limiting reagent.
3) The theoretical yield of C6H5CO2H is approximately 0.268 moles.

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)?

help me please