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14. Calculate the mass in grams of oxygen required to react completly with 0.75 moles of Aluminum.
4Al + 302 → 2Al2O3
a. 32 g of Oxygen
b. 18 g of Oxygen
c. 16 g of Oxygen
d. 9.0 g of Oxygen

Sagot :

  • 3mols of O_2 react with 4mol of Al
  • 3/4=0.75mol O_2 react with 1mol of Al

Moles of O_2:-

  • 0.75(0.75)
  • 0.5625mol

Mass of O_2:-

  • moles(Molar mass)
  • 0.5625(32)
  • 18g

Option B

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