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When the amount of oxygen is limited, carbon and oxygen react to form carbon monoxide. How many grams of CO can be formed from 35.0 grams of oxygen? 2C + O2 → 2CO Using 32.00 g/mole as the molecular mass of oxygen and 28.01 g/mole as the molecular mass of carbon monoxide, solve the above problem.

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JoshEast
The formula that denotes the incomplete combustion of carbon in the presence of a limited amount of oxygen is       2 C   +   O₂   →    2 CO

If the mass of oxygen that is used to combust carbon is 35 g
then the moles of oxygen = mass of oxygen ÷ molar mass of oxygen
                                           
                                            =  35 g  ÷  32 g/mol

                                            =  1.0938 mol

Now, the mole ratio of oxygen : carbon monoxide based on the balance equation is 1  :  2

⇒  If the mole of oxygen = 1.0938 mol
then the mole of carbon monoxide =  1.0938  ×  2
                                                           =  2.1876 mol

Mass of CO is =  mol of CO × molar mass of CO
                        =  2.1876 mol  ×  28 g/mol
                        =  61.25 g

∴ when a certain mass of carbon is combusted in 35g of oxygen then it produces approx. 61.25 g of Carbon Monoxide.  
CintiaSalazar

Considering the reaction stoichiometry, the mass of CO produced is 61.27 grams.

The balanced reaction is:

2 C + O₂ → 2 CO  

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • C: 2 moles
  • O₂: 1  mole
  • CO: 2 moles

The molar mass of each compound is:

  • C: 12.01 [tex]\frac{g}{mole}[/tex]
  • O₂: 32 [tex]\frac{g}{mole}[/tex]
  • CO: 28.01 [tex]\frac{g}{mole}[/tex]

Then, by reaction stoichiometry, the following amount of mass of each compound participate in the reaction:

  • C: 2 moles× 12.01 [tex]\frac{g}{mole}[/tex]= 24.02 grams
  • O₂: 1 mole× 32 [tex]\frac{g}{mole}[/tex]= 32 grams
  • CO: 2 moles× 28.01 [tex]\frac{g}{mole}[/tex]= 56.02 grams

Then you can apply the following rule of three:  if by stoichiometry 32 grams of oxygen form 56.02 grams of CO, 35 grams of oxygen form how much mass of CO?

[tex]mass of CO=\frac{35 grams of oxygenx56.02 grams of CO}{32 grams of oxygen}[/tex]

mass of CO= 61.27 grams

Finally, the mass of CO produced is 61.27 grams.

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