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3 years ago

2 FeCl2

1 answer:

4 0

Taking into account the reaction stoichiometry and limiting reagent, 0.7977 grams of KClO₃ are formed from the reaction of 2.50 g of KClO₄ with 150 mg of FeCl₂ in excess of H₂SO₄.

In first place, the balanced reaction is:

2 FeCl₂ + 13 KClO₄ + H₂SO₄ → 2 Fe(ClO₃)₃ + K₂SO₄ + 11 KClO₃ + H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

FeCl₂: 2 moles
KClO₄: 13 moles
H₂SO₄: 1 mole
Fe(ClO₃)₃: 2 moles
K₂SO₄: 1 mole
KClO₃: 11 moles
H₂O: 1 mole

The molar mass of the compounds is:

FeCl₂: 126.75 g/mole
KClO₄: 138.55 g/mole
H₂SO₄: 98 g/mole
Fe(ClO₃)₃: 306.2 g/mole
K₂SO₄: 174.2 g/mole
KClO₃: 122.55 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

FeCl₂: 2 moles× 126.75 g/mole= 253.5 grams
KClO₄: 13 moles× 138.55 g/mol= 1801.15 grams
H₂SO₄: 1 mole× 98 g/mole= 98 grams
Fe(ClO₃)₃: 2 moles× 306.2 g/mole= 612.4 grams
K₂SO₄: 1 mole× 174.2 g/mole= 174.2 grams
KClO₃: 11 moles× 122.55 g/mole= 1348.05 grams
H₂O: 1 mole× 18 g/mole= 18 grams

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use the reaction stoichiometry of the reaction and a simple rule of three as follows: if by stoichiometry 1801.15 grams of KClO₄ reacts with 253.5 grams of FeCl₂, if 2.50 grams of KClO₄ react how many mass of FeCl₂ will be needed?

$mass of FeCl_{2}=\frac{2.50 grams of KClO_{4} x253.5 grams of FeCl_{2}}{1801.15 grams grams of KClO_{4}}$

mass of FeCl₂=0.35 grams

But 0.35 grams of FeCl₂ are not available, 150 mg= 0.150 grams (being 1000 mg= 1 grams) are available. Since you have less mass than you need to react with 1801.15 grams of KClO₄, FeCl₂ will be the limiting reagent.

Then the following rule of three can be applied: if by reaction stoichiometry 253.5 grams of FeCl₂ form 1348.05 grams of KClO₃, 0.150 grams of FeCl₂ form how much mass of KClO₃?

$mass of KClO_{3}=\frac{0.150 grams of FeCl_{2}x1348.05 grams of KClO_{3} }{253.5grams of FeCl_{2} }$

mass of KClO₃= 0.7977 grams

Then, 0.7977 grams of KClO₃ are formed from the reaction of 2.50 g of KClO₄ with 150 mg of FeCl₂ in excess of H₂SO₄.

Learn more about reaction stoichiometry:

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Answer:

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Explanation:

Its a combustion reaction and they are always exothermic in nature.

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3 years ago

How many moles are in 2.04 x 10^8 atoms of calcium?

I am Lyosha [343]

Answer:

2.0 moles

Explanation:

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3 years ago

Coal containing 15.0% H2O, 2.0% S and 83.0% C by mass is burnt with the stoichiometric amount of air in a furnace. What is the m

devlian [24]

Answer:

This is a coal combustion process and we will assume

Inlet coal amount = 100kg

It means that there are

15kg of H2O, 2kg of Sulphur and 83kg of Carbon

Now to find the mole fraction of SO2(g) in the exhaust?

Molar mass of S = 32kg/kmol

Initial moles n of S = 2/32 = 0.0625kmols

Reaction: S + O₂ = SO₂

That is 1 mole of S reacts with 1 mole of O₂ to give 1 mole of SO₂

Then, it means for 0.0625 kmoles of S, we will have 0.0625 kmole of SO2 coming out of the exhaust

The mole fraction of SO2(g) in the exhaust=0.0625kmols

Explanation:

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3 years ago

A 7.00-mL portion of 8.00 M stock solution is to be diluted to 0.800 M. What will be the final volume after dilution? Enter your

amm1812

Explanation:

The number of moles of solute present in liter of solution is defined as molarity.

Mathematically, Molarity = $\frac{\text{no. of moles}}{\text{Volume in liter}}$

Also, when number of moles are equal in a solution then the formula will be as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

It is given that $M_{1}$ is 8.00 M, $V_{1}$ is 7.00 mL, and $M_{2}$ is 0.80 M.

Hence, calculate the value of $V_{2}$ using above formula as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

$8.00 M \times 7.00 mL = 0.80 M \times V_{2}$

$V_{2} = \frac{56 M. mL}{0.80 M}$

= 70 ml

Thus, we can conclude that the volume after dilution is 70 ml.

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3 years ago

A container with a fixed volume, filled with hydrogen gas at -104°C and 71.8 K PA is heated until the pressure reaches 225.9 K P

Daniel [21]

Answer:

The correct answer is 532 K

Explanation:

The Gay-Lussac law describes the behavior of a gas at constant volume, by changing the pressure or temperature. When is heated, the change in pressure of the gas is directly proportional to it absolute temperature (in Kelvin or K).

We have the following initial conditions:

P1= 71.8 kPa

T1= -104ºC +273 = 169 K

If the pressure increases until reaching 225.9 kPa (P2), we can calculate the final temperature of the gas (T2) by using the Gay-Lussac derived expression:

P1 x T2 = P2 x T1

⇒T2= (P2 x T1)/P1 = (225.9 kPa x 169 K)/71.8 kPa= 531.7 K ≅ 532 K

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4 years ago