If 50.0 g of KCl reacts with 50.0 g of O2 to produce KClO3 according to the following equation, how many grams of KClO3 will be formed? Word and Formula equation.

If 50.0 G Of KCl Reacts With 50.0 G Of O2 To Produce KClO3 According To The Following Equation, How Many

Answers

Answer 1

Answer:

A. 82.2g of KClO3

B. Word equation:

50g of KCl react with 50g of O2 to produce 82.2g of KClO3

C. Formula equation:

2KCl + 3O2 —> 2KClO3

Explanation:

The balanced equation for the reaction. This is given below:

2KCl + 3O2 —> 2KClO3

Next, we shall determine the masses of KCl and O2 that reacted and the mass of KClO3 produced from the balanced equation. This is illustrated below:

Molar Mass of KCl = 39 + 35.5 = 74.5g/mol

Mass of KCl from the balanced equation = 2 x 74.5 = 149g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 122.5g/mol

Mass of KClO3 from the balanced equation = 2 x 122.5 = 245g

Summary:

From the balanced equation above:

149g of KCl reacted.

96g of O2 reacted.

245g of KCl were produced.

Next, we shall determine the limiting reactant. This is illustrated below:

From the balanced equation above,

149g of KCl reacted with 96g of O2.

Therefore, 50g of KCl will react with = (50 x 96)/149 = 32.21g of O2.

Since a lesser mass of O2 ( i.e 32.21g) than what was given (i.e 50g) is needed to react completely with 50g of KCl, therefore, KCl is the limiting reactant and O2 is the excess reactant.

A. Determination of the mass of KClO3 produced from the reaction.

In this case the limiting reactant will be used.

From the balanced equation above,

149g of KCl reacted To produce 245g of KClO3.

Therefore, 50g of KCl will react to produce = (50 x 245)/149 = 82.2g of KClO3.

Therefore, 82.2g of KClO3 is produced from the reaction.

B. Word equation:

50g of KCl react with 50g of O2 to produce 82.2g of KClO3.

C. Formula equation:

2KCl + 3O2 —> 2KClO3

Answer 2

The mass of [tex]\rm KClO_3[/tex] produced has been 82.1085 g.

The word equation for the reaction has been: 50 gram potassium chloride reacts with 50 g oxygen to give 82.1085 g potassium chlorate.

The formula equation has been, [tex]\rm 2\;KCl\;+\;3\;O_2\;\rightarrow\;2\;KClO_3[/tex].

The balanced chemical equation for the reaction has been given as:

[tex]\rm 2\;KCl\;+\;3\;O_2\;\rightarrow\;2\;KClO_3[/tex]

Computation for Mass of Potassium chlorate:

From the balanced chemical equation, 2 moles of KCl reacts with 3 moles of oxygen to give 2 moles of [tex]\rm KClO_3[/tex].

The moles of reactants have been given as:

[tex]\rm Moles=\dfrac{mass}{molar\;mass}[/tex]

The moles of 50 g KCl has been given as:

[tex]\rm Moles\;KCl=\dfrac{50}{74.55} \\Moles\;KCl=0.67\;mol[/tex]

The moles of KCl available has been 0.67 mol.

The moles of 50 g [tex]\rm O_2[/tex] has been given as:

[tex]\rm Moles\;O_2=\dfrac{50}{32}\\Moles\;O_2=1.5625\;mol[/tex]

The moles of 50 g [tex]\rm O_2[/tex] has been 1.5625 mol.

From the balanced equation, for 2 moles KCl, 3 moles oxygen has been required. For 0.67 mol KCl, oxygen required has been:

[tex]\rm 2\;mole\;KCl=3\;moles\;O_2\\0.67\;mol\;KCl=\dfrac{3}{2}\;\times\;0.67\;mol\;O_2\\0.67\;mol\;KCl=1.005\;mol\;O_2[/tex]

The available moles of oxygen has been 1.5625 mol. Thus, oxygen has been excess reactant and KCl has been limiting reactant.

The moles of [tex]\rm KClO_3[/tex] produced has been given as:

[tex]\rm 2\;mol\;KCl=2\;mol\;KClO_3\\0.67\;mol\;KCl=\dfrac{2}{2}\;\times\;0.67\;mol\;KClO_3\\0.67\;mol\;KCl=0.67\;mol\;KClO_3[/tex]

The moles of [tex]\rm KClO_3[/tex] produced has been 0.67 mol.

The mass of [tex]\rm KClO_3[/tex] produced has been:

[tex]\rm Mass=Moles\;\times\;molar\;mass\\Mass\;KCl=0.67\;\times\;122.55\;\\Mass\;of\;KCl=82.1085\;g[/tex]

The mass of [tex]\rm KClO_3[/tex] produced has been 82.1085 g.

The word equation for the reaction has been:

50 gram potassium chloride reacts with 50 g oxygen to give 82.1085 g potassium chlorate.

The formula equation has been, [tex]\rm 2\;KCl\;+\;3\;O_2\;\rightarrow\;2\;KClO_3[/tex].

For more information about chemical equation, refer to the link:

https://brainly.com/question/13350862


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Answers

Hey there!

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CO+3H2⇌CH4+H2O



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Answers

Answer:

The correct answer is  

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Answers

Answer:

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Explanation:

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Explanation:

Answer:

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Answers

Answer:

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Explanation:

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Answers

Answer:

heat = 229139.43 cal

Explanation:

given data

specific heat of iron is 0.107 cal/g-°C     C = 0.4494 J/g°C;

Mass of iron = 24.7 kg = 24700 g

temperature T1 = 880°C

temperature T2 = 13°C

solution

we know that Heat lost that is express as

heat loss = mcΔT      .........................1

here m is the mass and c is specific heat capacity of iron and ΔT is the change in temperature

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Answers

Answer:

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Answers

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Answer:

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Answers

Answer:

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What is the percent yield of C when 35 g CS₂ are reacted and produce 36.4 g C. * LABEL THE LIMITING REACTANT IN YOUR WORK.* Use the following balanced chemical equation to solve the problem: CS₂ + 4 CO --> 5 C + 2 SO₂ *

Answers

Answer:

Limiting reactant is CS2

% yield = 131.8 %

Explanation:

Using the alanced equation given, Mole ratio of C to CS2 is 1 : 5.

That is, for every 1 mole of CS2 used, 5 moles of Carbon is produced.

35g of CS2 --- [tex]\frac{35}{76.14}[/tex] moles

= 0.4597 moles

⇒ mass of C produced = 5 × 0.4597 × 12 g/mol

= 27.6 g of C.

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Answers

Answer:

C

Explanation:

The answer is always C

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Explanation:

When the surface of a mirror curves inward, like the inside of a bowl, it is called c. Concave mirror.

ASAP NEED HELP PLSSS
A gas occupying a volume of 3.75L at a pressure of 0.980 arm is allowed to expand at constant temperature until its pressure reaches 0.641 atm. What is the final volume of the sample?

Answers

Answer:

Volume = 5.73L

Explanation:

Data;

V1 = 3.75L

P1 = 0.980atm

P2 = 0.641atm

V2 = ?

This question involves the use of Boyle's law, which states that, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.

Mathematically,

V = kP, k = PV

P1V1 = P2V2 =P3V3=........=PnVn

P1V1 = P2V2

V2 = (P1 * V1) / P2

V2 = (0.980 × 3.75) / 0.641

V2 = 5.73L

The final volume of the gas is 5.73L

the combustion of 0.1240 kg of propane in the presence of excess oxygen produces 0.3110kg of carbon dioxide what is the limittinf reacrant?

Answers

Answer:

Propane

Explanation:

From the question given, we were told that 0.1240 kg of propane reacted with excess oxygen to produce 0.3110kg of carbon dioxide.

Since the reaction took place in the presence of excess oxygen, therefore, propane is the limiting reactant as all of it is used up in the presence of excess oxygen.

Answer:

The limiting reactant is propane, C₃H₈ and the percentage yield is 83.77%

Explanation:

Mass of propane = 0.1240 kg = 124 g

Mass of carbon dioxide = 0.3110 kg = 311 g

Molar mass of propane = 44.1 g/mol

Molar mass of carbon dioxide = 44.01 g/mol

Number of moles of propane = 124/44.1 = 2.812 moles

Number of moles of carbon dioxide = 311/44.01 = 7.067 moles

Equation for the reaction

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Hence 1 mole of propane ideally yields 3 moles of CO₂

Hence, 2.812 moles of propane will yield 3×2.812 moles = 8.44 moles of CO₂

Since, oxygen is in excess, therefore, the limiting reactant = Propane, C₃H₈

The percentage yield = 7.067/8.44× 100 = 83.77%.

How much NaOH (in grams) is needed to prepare 463 mL of solution with a pH of 10.020?

Answers

Answer: [tex]1.94\times 10^{-3}g[/tex] of NaOH

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

[tex]pH=-\log [H^+][/tex]

Putting in the values:

[tex]10.020=-\log[H^+][/tex]

[tex][H^+]=9.55\times 10^{-11}[/tex]

[tex][H^+][OH^-]=10^{-14}[/tex]

[tex][OH^-]=\frac{10^{-14}}{9.55\times 10^{-11}}=1.05\times 10^{-4}M[/tex]

[tex]NaOH\rightarrow Na^++OH^-[/tex]

[tex]Molarity=\frac{moles\times 1000}{\text {Volume in ml}}[/tex]

[tex]1.05\times 10^{-4}M=\frac{moles\times 1000}{463ml}[/tex]

moles = [tex]4.86\times 10^{-5}[/tex]

Mass of [tex]NaOH=moles\times {\text {Molar mass}}=4.86\times 10^{-5}\times 40=1.94\times 10^{-3}g[/tex]

Thus [tex]1.94\times 10^{-3}g[/tex] of NaOH is needed to prepare 463 mL of solution with a pH of 10.020


(b) Which peak, the one at 365 nm or the one at 435 nm, results from the absorption of
photons with the greater energy? Justify your answer.
(c) Based on the diagram above, is the forward reaction endothermic or exothermic? Justify
your answer in terms of Le Châtelier's principle.

Answers

Answer:

365nm and Exothermic reaction

Explanation:

b) The Einstein Planck's equation states that

E = hf where E is the energy of a photon

                     h  is the Planck's constant

                     f  is the frequency

But Frequency f = c/λ  where c  is the speed of light

                                                  λ  is the wavelength

Therefore Energy E = hc/λ

From the above equation we see that the Energy of a photon is indirectly proportional to its wavelength

Hence as the wavelength increases from 365nm to 435nm, the energy of the photons decreases

Therefore, a photon will wavelength 365nm would have a higher energy than one will wavelength of 435nm.

c) From the diagram we observe the following:

At high temperatures,   HPR⁺ occurs at a higher concentration than at low temperatures.

At low temperatures, more of PR⁻ is formed than at high temperatures

The equation of the reaction is

HPR⁺ (aq) ⁺ ⇆ H⁺ + PR⁻(aq)      

Le Chartelier's principle states that when an external constraint such as a change in concentration, pressure or temperature is placed on a system in equilibrium, the system acts in such a way as to oppose the change.

Hence a decrease in temperature on the side of reactant, HPR would force the equilibrium to the side of the products, PR in other to accommodate the change

The forward reaction, which produces more of the product PR⁻ is exothermic because  the at low temperatures, the the forward reaction is favoured and equilibrium shifts to the right.

Katie throws a 3 kilogram ball fast at a speed of 15 meters what is the momentum of the ball katie threw

Answers

Answer:

Hey!

Your answer is 45kg.m/s

Explanation:

Using the formula P=M x V

(p=momentum...M=mass...V=velocity)

We do M x V

15 x 3

Which gives us an answer of 45kg.m/s!

HOPE THIS HELPS!!


Which is usually associated with a slower reaction rate?
A. using a whole solid in the reactants
B. adding heat energy to the reactants
C. adding food coloring to the reactants
D. using a strong solution of the reactants

Answers

A. Using a whole solid in the reactants
using a whole solid in the reactants

Relate dark matter to the development of the universe after the Big Bang. In 3-5 sentences, speculate on how the development of the universe would have been different if there had been no dark matter. In your answer, use the term structures.

Please answer this and ill mark you as brainliest

Answers

Answer:

Dark matter also called baronic matter is the matter that makes up 27% of the universe.

In 1933 it was determined as that mass that cannot be seen, that is, the non-visible mass of outer space.

Dark matter also plays a central role in the formation of structures and the evolution of galaxies and has measurable effects on the anisotropy of cosmic microwave background radiation. The composition of this matter is unknown today.

The dark matter component has considerably more mass than the "visible" component of the Universe.

Explanation:

There are certain researchers who say that the appearance of dark matter was before the appearance of the big bang.

A relevant fact of this matter is that dark matter exerts gravity, and that gravity affects the movements of objects.

Despite the fact that nothing is known about its origin, astronomers have amply demonstrated that dark matter plays a determining role in the formation of galaxies and galactic clusters, which could not maintain their cohesion without its existence, but many doubt that it is the remainder / remnant or product of a big bang.

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