Pro∆G0=-RT in k Known as
1 answer:
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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:
P = 0.6815 atm
Explanation:
Pressure = 754 torr
The conversion of P(torr) to P(atm) is shown below:
So,
Pressure = 754 / 760 atm = 0.9921 atm
Temperature = 294 K
Volume = 3.1 L
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9921 atm × 3.1 L = n × 0.0821 L.atm/K.mol × 294 K
⇒n of helium gas= 0.1274 moles
Surface are = 1257 cm²
For a sphere, Surface area = 4 × π × r² = 1257 cm²
r² = 1257 / 4 × π ≅ 100 cm²
r = 10 cm
The volume of the sphere is :
Where, V is the volume
r is the radius
V = 4190.4762 cm³
1 cm³ = 0.001 L
So, V (max) = 4.19 L
T = 273 K
n = 0.1274 moles
Using ideal gas equation as:
PV=nRT
Applying the equation as:
P × 4.19 L = 0.1274 × 0.0821 L.atm/K.mol × 273 K
<u>P = 0.6815 atm</u>
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