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How many grams of carbon dioxide will be produced if 76.4 grams of
C2H3Br3 reacted with 49.1 grams of O2? How many grams of the
excess reactant will be left after the reaction is over? Consider the
equation: C2H3Br3+O2 -> CO2+H2O+Br2

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Answer:

Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g × (1 mole/ 32 g) = 1.53 moles O2 B.

Explanation:

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https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield
  • Step:1  When we are Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556  with the moles of C2H3Br3 are required. Just Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.
  • Step:2 When there are Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: when is equal to
  • Conclusion is that: Thus, that A. 76.4g × (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g × (1 mole/ 32 g) = 1.53 moles O2 B.

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