Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g
1 answer:
Answer:
156.4g K
Explanation:
I'm not sure if it is correct but I think it should be this
What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2
What are we looking for?: #g of K
What is the ratio of K to Cl2?: 2:1
Set up equation: 2molCl2 x
Cancel unwanted units: 2 x
Answer we got: 2 x 2mol K = 4mol K
Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K
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Answer:
a) The theoretical yield is 408.45g of
b) Percent yield =
Explanation:
1. First determine the numer of moles of and
.
Molarity is expressed as:
M=
- For the
M=
Therefore there are 1.75 moles of
- For the
M=}{1Lsolution}[/tex]
Therefore there are 2.0 moles of
2. Write the balanced chemical equation for the synthesis of the barium white pigment, :
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the :
- For the :
As the is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield =
Percent yield =
The real yield is the quantity of barium white pigment you obtained in the laboratory.