Answer:
- <u><em>O₂ is the limiting reactant.</em></u>
Explanation:
To find which of the <em>available reactants</em> is<em> limiting</em> the reaction, you need to compare the ratio of the available reactants with the stoichiometric mole ratio from the chemical equation.
The balanced chemical equation is:
The coefficients of each reactant indicate the stoichiometric mole: 4 moles of Al react with3 moles of O₂.
To find the ratio of the available reactants convert the masses in grams to number of moles.
- number of moles = mass in grams / molar mass
The atomic mass of Al is 26.982g/mol
- number of moles of Al = 3.17g / 26.982g/mol = 0.1175 mol
The molar mass of O₂ is 31.998g/mol
- number of moles of O₂ = 2.55g / 31.998g/mol = 0.07969 mol
<u>Theoretical ratio </u>
<u>Ratio of available reactants</u>
<u>Comparison</u>
Since the ratio of the available amount of Al per mol of available O₂ is greater than the theoretical ratio of Al per mol of O₂, there is more available Al than what is needed to react. That implies that when all the O₂ is consumed, there will be some Al left. O₂ is limiting ther reaction.
The reactant that is consumed completely is the limiting reactant, and the reactant that is left is the excess reactant; thus, O₂ is the limiting reactant.
The idea behind a tree diagram is to start on the left with the whole thing, or one. Every time several possible outcomes exist the probability in that branch splits off into a smaller branch for each outcome.
Answer:
Heyy Mason, I donʻt know if you got the answer already but I was also trying to find the answer haha and came across this. Might as well help you since I eventually stopped being lazy and did the calculations...The answer is 144 L.
Explanation:
Use Charles Law to solve this
V1 = 50.0L
T1 = 205K
V2 = ?
T2 = 590K
using V1/T1 = V2/T2 plug in the numbers
50.0L/205K = V2/590K
Switch it around so V2 is isolated
V2 = 590K(50.0L/205K)
cancel out the K
solve
you get 143.90243
round to 144
C. A triangle will generate the most