The initial pressure of a gas is 1.58 atm and occupies 1.76 L of space at constant temperature. This gas is compressed so that i
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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
<u>Answer:</u> The mass of precipitate (lead (II) iodide) that will form is 119.89 grams
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
Molarity of NaI solution = 0.130 M
Volume of solution = 0.400 L
Putting values in above equation, we get:
The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:
The precipitate (insoluble salt) formed is lead (II) iodide
By Stoichiometry of the reaction:
2 moles of NaI produces 1 mole of lead (II) iodide
So, 0.52 moles of NaI will produce = of lead (II) iodide
To calculate the number of moles, we use the equation:
Moles of lead (II) iodide = 0.26 moles
Molar mass of lead (II) iodide = 461.1 g/mol
Putting values in above equation, we get:
Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams