consider the reaction between 60ml of liquid methanol, ch3OH, (density=0.850 g/m

Question

consider the reaction between 60ml of liquid methanol, ch3OH, (density=0.850 g/ml) and 30 L of O2 at 30 degrees Celsius and a pressure of 2atm. If the reaction goes to completion calculate the number of moles of CO2 and water formed. consider the reaction between 60ml of liquid methanol, ch3OH, (density=0.850 g/ml) and 30 L of O2 at 30 degrees Celsius and a pressure of 2atm. If the reaction goes to completion calculate the number of moles of CO2 and water formed. consider the reaction between 60ml of liquid methanol, ch3OH, (density=0.850 g/ml) and 30 L of O2 at 30 degrees Celsius and a pressure of 2atm. If the reaction goes to completion calculate the number of moles of CO2 and water formed.

Explanation / Answer

The Reaction is

2CH3OH + 3O2 ----> 2CO2 + 4H2O

First we have to find the Limiting Reagent. For that we have to find the number of moles of each reactant that is being used.

1) Number of moles of CH3OH = mass/ Molecular weight
   Mass = Volume * Density
   = 60 mL * 0.85 g/mL
= 51 g
Molecular mass of the CH3OH = 32 g
  moles = 51/32 = 1.593 moles

2) Moles of Oxygen, this can be found out using the Ideal Gas Equation. ( Oxygen is assumed as Ideal Gas)

   PV = nRT
  Given P = 2 atm, V = 30L and T = 30 + 273 = 303K
R (Universal Gas constant ) = 0.0821 L.atm/(mol.K)

Using the values of the parameters in the ideal gas equation gives us
   2* 30 = n * 0.0821*303
  n = 2.411 moles

For 2 moles of CH3OH, number of moles of O2 required is 3
So for 1.593 moles of CH3OH, moles of O2 required = (3/2)*1.593 = 2.3895

Hence the limiting Reagent is CH3OH

The number of moles of CO2 formed = ( Coefficient of CO2 / Coefficient of CH3OH )* moles of CH3OH
   = (2/2)*1.593
   = 1.593 moles of CO2 is formed
     
  
   

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