Question

2 FeCl2

Answer 1

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:

• brainly.com/question/24741074
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• brainly.com/question/23871710