Hello!
1) In moles
The chemical reaction between para-xylene and oxygen to produce terephthalic acid and water is the following one:
C₆H₄(CH₃)₂ + 3O₂ → C₆H₄(CO₂H)₂ + 2H₂O
To calculate how much terephthalic acid in moles could be made from 154 grams of para-xylene in moles, we need to apply the following conversion factor to go from grams of para-xylene to moles of terephthalic acid, using the molar mass and the reaction coefficients:
[tex]154gC_6H_4(CH_3)_2* \frac{1 mol_{C_6H_4(CH_3)_2}}{106,16 g_{C_6H_4(CH_3)_2}}* \frac{1 mol _{C_6H_4(CO_2H)_2}}{1 mol_{C_6H_4(CH_3)_2}} \\ =1,4506moles_{C_6H_4(CO_2H)_2} [/tex]
So, 1,4506 moles of Terephthalic Acid can be produced from 154 grams of para-xylene.
2) In grams
The process is similar as above. The chemical reaction between para-xylene and oxygen to produce terephthalic acid and water is the following one:
C₆H₄(CH₃)₂ + 3O₂ → C₆H₄(CO₂H)₂ + 2H₂O
To calculate how much terephthalic acid in grams could be made from 154 grams of para-xylene in moles, we need to apply the following conversion factor (similar as above) to go from grams of para-xylene to grams of terephthalic acid, using the molar masses of both compounds and the reaction coefficients:
[tex]154gC_6H_4(CH_3)_2* \frac{1 mol_{C_6H_4(CH_3)_2}}{106,16 g_{C_6H_4(CH_3)_2}}* \frac{1 mol _{C_6H_4(CO_2H)_2}}{1 mol_{C_6H_4(CH_3)_2}}* \frac{166,13 g}{1 mol _{C_6H_4(CO_2H)_2}} \\ =240,995g_{C_6H_4(CO_2H)_2} [/tex]
The process is almost the same as above, just adding the calculation of the grams of terephthalic acids from the moles.
So, 240,995 grams of Terephthalic Acid can be produced from 154 grams of para-xylene.
Have a nice day!
