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kotykmax [81]
2 years ago
14

If 5.0 g of each reactant were used for the the following process, the limiting reactant would be: 2KMnO4 + 5Hg2Cl2 + 16HCl -&gt

; 10HgCl2 + 2MnCl2 + 2KCl + 8H2O (KMnO4 MM=158 g/mol; Hg2Cl2 MM=472.1 g/mol; HCl MM=36.5 g/mol; HgCl2 MM=271.5 g/mol; MnCl2 MM=125.8 g/mol; KCl MM=74.6 g/mol; H2O MM=18 g/mol) *
Chemistry
1 answer:
IceJOKER [234]2 years ago
7 0

Answer:

The limiting reactant is Hg2Cl2.

Explanation:

Step 1: Data given

Mass of each reactant = 5.0 grams

KMnO4 MM=158 g/mol

Hg2Cl2 MM=472.1 g/mol

HCl MM=36.5 g/mol

HgCl2 MM=271.5 g/mol

MnCl2 MM=125.8 g/mol

KCl MM=74.6 g/mol

H2O MM=18 g/mol)

Step 2: The balanced equation

2KMnO4 + 5Hg2Cl2 + 16HCl -> 10HgCl2 + 2MnCl2 + 2KCl + 8H2O

Step 3: Calculate moles

KMnO4 = 5.00 grams / 158 g/mol = 0.0316 mol

Hg2CL2 = 5.00 grams / 472.1 g/mol = 0.0106 mol

HCl = 5.00 grams / 36.5 g/mol = 0.137 mol

Step 3: Calculate limiting reactant

For 2 moles of KMno4 we need 5 moles of Hg2Cl2 and 16 moles of HCl

Hg2Cl2 has the smallest amount of moles.

For 5 moles Hg2Cl2 ( 0.0106 mol) we need 0.0106 / (5/2) = 0.00424 mol KMnO4

For 5 moles Hg2Cl2 we need (16/5) *0.0106 = 0.03392 moles of HCl

So the limiting reactant is Hg2Cl2.

Step 4: Calculate moles of product produced:

2*0.0106 = 0.0212 moles of HgCl2

(2/5) * 0.0106 = 0.00424 moles of MnCl2 and 0.00424 moles of KCl

(8/5) * 0.0106 = 0.01696 moles H2O

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Answer: A

Explanation:

Considering the order of filling electrons into orbitals, movement from a higher to a lower energy level results in the emission of a photon with energy equal to the energy difference between the two energy levels. However, the energies of different orbitals are close together for high values of n (principal quantum number). Their relative energies may change significantly when they form ions. This implies that energy levels are better separated and have high differences in energy for low values of n. Hence the answer. This means that photons transiting between these Lowe n levels will posses higher photon energy due to larger energy difference between levels.

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2 years ago
What was the original volume of a gas if heating it from 22k to 85k produced a volume of 3.8 liters?
mel-nik [20]

Answer: the original volume was 0.98L

Explanation:

V1 =?

T1 = 22k

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V1 /T1 = V2 /T2

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V1 = 22 x (3.8/85)

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3 0
2 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ →  3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

  • Molar mass(MM) of Cr₂(SO₄)₃:

        = 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

        = 2*(52) + 3*(32) + 12*(16)

        = 104 + 96 + 192

        = 392 g

  • Molar mass(MM) of K₂SO₄:

        = 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

        = 2*(39) + 32 + 4*(16)

        = 78 + 32 + 64

        = 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : \frac{522 * 450}{392}

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0
2 years ago
Directions: Select the choice that best fits each statement. The following questions(s) refer to the following elements. (A) Car
Illusion [34]

Answer:

sulfur

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During mining, sulfur from pyrite contributes to acid-mine drainage.

3 0
2 years ago
Cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture. The initial rea
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Answer : The correct option is, (C) ^{58}\textrm{Fe}

Explanation :

Neutron capture : In this decay process, an atomic nucleus and one or more number of neutrons collide and combine to form a heavier nucleus. The mass number changes in this process.

The neutron capture equation is represented as,

_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma

(A is the atomic mass number and Z is the atomic number)

Beta emission or beta minus decay : It is a type of decay process, in which a neutrons gets converted to proton, an electron and anti-neutrino. In this the atomic mass number remains same.

The beta minus decay equation is represented as,

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e

(A is the atomic mass number and Z is the atomic number)

As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma

Process 2 : Beta emission.

_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e

Process 3 : Neutron capture.

_{27}^{59}\textrm{Co}+_{0}^1\textrm{n}\rightarrow _{26}^{60}\textrm{Co}+\gamma

From this we conclude that, the initial reactant in the production of cobalt-60 is _{26}^{58}\textrm{Fe}

Hence, the correct option is, (C) ^{58}\textrm{Fe}

7 0
2 years ago
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