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Answer:

25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry, the following amounts of moles of each compound participate in the reaction:

  • C₂H₂: 2 moles
  • O₂: 5 moles
  • CO₂: 4 moles
  • H₂O: 2 moles

The molar mass of each compound is:

  • C₂H₂: 26 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

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

  • C₂H₂: 2 moles* 26 g/mole= 52 g
  • O₂: 5 moles* 32 g/mole= 160 g
  • CO₂: 4 moles* 44 g/mole= 176 g
  • H₂O: 2 moles* 18 g/mole= 36 g

Then you can apply the following rule of three: if by stoichiometry 52 grams of C₂H₂ react with 160 grams of O₂, 8.15 grams of C₂H₂ react with how much mass of O₂?

[tex]mass of O_{2} =\frac{8.15 grams of C_{2} H_{2}*160 grams of O_{2} }{52 grams of C_{2} H_{2}}[/tex]

mass of O₂= 25.08 grams

25.08 grams of O₂ are needed to react with 8.15 g of C₂H₂.

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