The 3.13 g of K would be needed to completely react with the remaining [tex]Cl_2[/tex].
To determine which reactant is limiting, we need to calculate the amount of product that can be formed from each reactant and compare them. The reactant that produces less product is the limiting reactant, since the reaction cannot proceed further once it is consumed.
The balanced chemical equation for the reaction between solid potassium and chlorine gas is:
2 K(s) + [tex]Cl_2[/tex](g) -> 2 KCl(s)
From the equation, we can see that 2 moles of K react with 1 mole of [tex]Cl_2[/tex] to form 2 moles of KCl.
First, we need to convert the masses of K and [tex]Cl_2[/tex] into moles:
moles of K = 15.20 g / 39.10 g/mol = 0.388 mol
moles of [tex]Cl_2[/tex] = 2.830 g / 70.90 g/mol = 0.040 mol
Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of KCl from each reactant:
Theoretical yield of KCl from K: 0.388 mol K x (2 mol KCl / 2 mol K) = 0.388 mol KCl
Theoretical yield of KCl from [tex]Cl_2[/tex]: 0.040 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) = 0.080 mol KCl
We can see that the theoretical yield of KCl from K is 0.388 mol, while the theoretical yield of KCl from [tex]Cl_2[/tex] is 0.080 mol. Therefore, the limiting reactant is [tex]Cl_2[/tex], since it produces less product.
To determine how much more of the limiting reactant would be needed to completely consume the excess reactant, we can use the stoichiometry of the balanced equation.
We know that 1 mole of [tex]Cl_2[/tex] reacts with 2 moles of K to produce 2 moles of KCl. Therefore, the amount of additional K needed to react with the remaining [tex]Cl_2[/tex] can be calculated as follows:
moles of K needed = 0.040 mol [tex]Cl_2[/tex] x (2 mol K / 1 mol [tex]Cl_2[/tex])
= 0.080 mol K
This means that 0.080 moles of K would be needed to completely consume the remaining [tex]Cl_2[/tex]. We can convert this to a mass by multiplying by the molar mass of K:
mass of K needed = 0.080 mol K x 39.10 g/mol
= 3.13 g K
Therefore, The 3.13 g of K would be needed to completely react with the remaining.
Example verification:
Suppose we had an additional 0.50 g of [tex]Cl_2[/tex] in the reaction. Would all of the K be consumed, or would there still be excess K?
Moles of additional [tex]Cl_2[/tex] = mass of [tex]Cl_2[/tex] / molar mass of [tex]Cl_2[/tex]
Moles of additional [tex]Cl_2[/tex] = 0.50 g / 70.90 g/mol
Moles of additional [tex]Cl_2[/tex] = 0.0070 mol
The theoretical yield of KCl that can be formed from the additional [tex]Cl_2[/tex] is:
0.0070 mol [tex]Cl_2[/tex] x (2 mol KCl / 1 mol [tex]Cl_2[/tex]) x (74.55 g KCl / 1 mol KCl) = 1.04 g KCl
Therefore, the total amount of KCl that can be formed from all of the [tex]Cl_2[/tex] is:
5.95 g + 1.04 g = 6.99 g
The amount of K that would be needed to completely consume all of the [tex]Cl_2[/tex].
Learn more about Solid Potassium at
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In calorimetry, energy is measured through heat transfer from one substance to
another. Which of the following is NOT a method of heat transfer?
Answer:
Refraction
Explanation:
PLEASE HELP!! ORGANIC CHEMISTRY
A sample of a diatonic gas is loaded into an evacuated bottle at STP. The 0.25 L bottle contains 1.76 grams of the unidentified gas. Calculate the molar mass of the gas. What is the identity of the diatomic gas?
Answer:
(a) 157.7 g
(b) 7.04 g/dm³
Explanation:
(a) From the question,
According to Avogadro's Law,
1 mole of every gas at STP occupies a volume of 22.4 dm³
But mass of 1 mole of the diatomic gas = molar mass of the gas.
This Implies that,
The molar mass of the gas at STP occupies a volume of 22.4 dm³
From the question,
If,
0.25 L bottle contain 1.76 g of the gas,
Therefore,
Molar mass of the gas = (1.76×22.4)/0.25
Molar mass of the gas = 157.7 g.
(b) Density of the gas = mass/volume
D = m/v
Given: m = 1.76 g, v = 0.25 L = 0.25 dm³
Therefore,
D = 1.76/0,25
D = 7.04 g/dm³
What is the limiting reactant in the following equation? How much Fe2O3 will be produced if 2.1 g of Fe reacts with 2.1 g of O2?
4 Fe + 3O2 —> 2Fe2O3
Answer:
Fe is limiting reactant and 3.00g of Fe2O3 will be produced
Explanation:
To solve this question we must convert the mass of each reactant to moles and, using the reaction we can find limiting reactant. With moles of limiting reactant we can find moles of Fe2O3 and its mass as follows:
Moles Fe -Molar mass: 55.845g/mol-
2.1g * (1mol / 55.845g) = 0.0376 moles
Moles O2 -Molar mass: 32g/mol-
2.1g * (1mol / 32g) = 0.0656 moles
For a complete reaction of 0.0656 moles of O2 are needed:
0.0656moles O2 * (4mol Fe / 3 mol O2) = 0.0875 moles Fe
As there are just 0.0376 moles,
Fe is limiting reactant
The mass of Fe2O3 is:
Moles:
0.0376 moles Fe* (2mol Fe2O3 / 4mol Fe) = 0.0188 moles Fe2O3
Mass:
0.0188 moles Fe2O3 * (159.69g / mol) =
3.00g of Fe2O3 will be produced
The equivalence point of a titration corresponds to which of the following?
O the point where equal volumes of acid and base have been used
O Equivalence point is another term for end point
All of the listed options are true
Equivalence point is defined as the point where the pH indicator changes color
O the point where the acid and base have been added in proper stoichiometric amounts
Answer:
E: the point where the acid and base have been added in proper stoichiometric amounts
Explanation:
Equivalence point in titration is simply the point where the amounts of acid and base used just sufficiently reacts chemically to cause neutralization whereas the endpoint is the point where the indicator of the titration changes colour.
The Equivalence point occurs before the endpoint.
Thus, option E is correct.