What is the mass in grams of oxygen gas that is required to produce 32.1 grams of maleic acid anhydride?
1 answer:
Considering the reaction stoichiometry, the mass of oxygen gas that is required to produce 32.1 grams of maleic acid anhydride is 47.17 grams.
The balanced reaction is:
2 C₆H₆(l) + 9 O₂(g)→ 2 C₄H₂O₃(s) + 4 CO₂(g) + 4 H₂O(g)
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:
- C₆H₆: 2 moles
- O₂: 9 moles
- C₄H₂O₃: 2 moles
- CO₂: 4 moles
- H₂O: 4 moles
The molar mass of each compound is:
- C₆H₆: 78 g/mole
- O₂: 32 g/mole
- C₄H₂O₃: 98 g/mole
- CO₂: 44 g/mole
- H₂O: 18 g/mole
By reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:
- C₆H₆: 2 moles× 78 g/mole= 156 grams
- O₂: 9 moles× 32 g/mole= 288 grams
- C₄H₂O₃: 2 moles× 98 g/mole= 196 grams
- CO₂: 4 moles× 44 g/mole= 176 grams
- H₂O: 4 moles× 18 g/mole= 72 grams
Then you can apply the following rule of three: If by stoichiometries 196 grams of C₄H₂O₃ are produced by 288 grams of O₂, 32.1 grams of C₄H₂O₃ are produced from how much mass of O₂?
<u><em>mass of O₂= 47.17 grams</em></u>
Finally, the mass of oxygen gas that is required to produce 32.1 grams of maleic acid anhydride is 47.17 grams.
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