Answer:
A. fluorine, 1.79 moles
Explanation:
Given parameters:
Mass of carbon = 87.7g
Mass of fluorine gas = 136g
Unknown:
The limiting reactant and the maximum amount of moles of carbon tetrafluoride that can be produced = ?
Solution:
Equation of the reaction:
C + 2F₂ → CF₄
let us find the number of the moles the given species;
Number of moles =
C; molar mass = 12;
Number of moles =
= 7.31moles
F; molar mass = 2(19) = 38g/mol
Number of moles =
= 3.58moles
So;
From the give reaction:
1 mole of C requires 2 moles of F₂
7.31 moles of C will then require 2 x 7.31 moles of F₂ = 14.62moles
But we have 3.58 moles of the F₂;
Therefore, the reactant in short supply is F₂ and it is the limiting reactant;
So;
2 moles of F₂ will produce mole of CF₄
3.58 moles of F₂ will then produce
= 1.79moles of CF₄
Answer:
The molar concentration of this solution is 0.0463 mol/L
Explanation:
Step 1 : Data given
Mass of a nonelectrolyte solute = 2.69 grams
Volume of water = 345 mL = 0.345 L
Temperature = 26.0°CC = 273 + 26 = 299 K
The osmotic pressure = 863 torr
⇒ 863torr /760 = 1.13553 atm
Step 2: Calculate the molar concentration of this solution
Π = i*M*R*T
⇒with Π = the osmotic pressure = 1.13553 atm
⇒with i = the van't Hoff factor of the nonelectrolyte solute = 1
⇒with M = the molar concentration = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 299 K
1.13553 atm = 1 * M * 0.08206 L*atm/mol*K * 299 K
M = 1.13553 / (0.08206*299)
M = 0.0463 mol/L
The molar concentration of this solution is 0.0463 mol/L
Option C: Sulfur Dioxide is the answer
Hope this helps
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