Do I need to solve 12.046 x 10^25 to get the answer
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:
MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
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
C the products are on the right side of the arrow
Answer:
The pressure of the gas is 224839.8 atm
Explanation:
As we know
PV = nRT
Substituting the given values, we get -
P * 1 L = 90.1 moles * 8.314 4621(75). J K−1 mol−1 * 300
P = 224839.8 atm
The pressure of the gas is 224839.8 atm
