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
The given blank can be filled with endothermic reactions.
The endothermic reactions or processes captivates energy from its surroundings, primarily, but will not always be in the form of heat. A reaction, which requires the input of energy that dissociates a bond will always be endothermic. The best examples of endothermic reactions are evaporation of the liquid, melting cubes, melting or solid salts, and others.
When the enthalpy change of the reaction is positive the reaction is endothermic
this means that the reaction is absorbing energy from the surrounding to the vessel
