If 25 mol of C8H18 are available, how many mol of CO2 can be produced
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:
184.62 ml
Explanation:
Let and
be the initial and
and
be the final pressure, volume, and temperature of the gas respectively.
Given that the pressure remains constant, so
...(i)
= 200 ml
K
K
From the ideal gas equation, pv=mRT
Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.
For the initial condition,
For the final condition,
Equating equation (i), and (ii)
[from equation (i)]
Putting all the given values, we have
Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.