Answer:
The answers are options b , c and d.
Hope this helps.
It should be 24g of carbon
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
To get the percent yield, we will use this formula:
((Actual Yield)/(Theoretical Yield)) * 100%
Values given: actual yield is 220.0 g
theoretical yield is 275.6 g
Now, let us substitute the values given.
(220.0 grams)/(275.6 grams) = 0.7983
Then, to get the percentage, multiply the quotient by 100.
0.7983 (100) = 79.83%
Among the choices, the most plausible answer is 79.8%
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The answer is (3) Cu2O. Copper (I) has an oxidation state of +1 (that's what the "I" indicates). You can also think of this as copper (I) having a charge of +1. Oxygen has an oxidation state of -2 (that's just a rule you have to know), and you can think of it as oxygen having a charge of -2. You need oxidation numbers in a neutral compound to add up to 0 (or charges in a neutral compond to add up to 0), so you need two Cu to balance the O, which is Cu2O.