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zavuch27 [327]
2 years ago
7

What is the mass percent of potassium sulfate in solution if 78g of potassium sulfate is dissolved in 500 mL of water? (The dens

ity of water is 1.00 g/mL)
Chemistry
1 answer:
Debora [2.8K]2 years ago
6 0

Answer:

13.5 %

Explanation:

First we<u> calculate the mass of 500 mL of water</u>, using <em>its density</em>:

  • Volume * Density = Mass
  • 500 mL * 1.00 g/mL = 500 g

Then we <u>calculate the mass percent of potassium sulfate</u>, using the formula:

Mass of Potassium Sulfate / Total Mass * 100%

  • 78 g / (78 + 500) g * 100 % = 13.5 %

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In the chemical reaction:
velikii [3]

Taking into account the reaction stoichiometry and definition of limiting reactant, AgNO₃ will be the limiting reagent.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Cu + 2 AgNO₃  → 2 Ag + Cu(NO₃)₂

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:

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  • AgNO₃: 2 moles
  • Ag: 2 moles
  • Cu(NO₃)₂: 1 mole

<h3>Limiting reagent</h3>

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.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 1 mole of Cu reacts with 2 moles of AgNO₃, 1.8 moles of Cu reacts with how many moles of AgNO₃?

amount of moles of AgNO_{3} =\frac{1.8 moles of Cux2 moles of AgNO_{3} }{1 mole of Cu}

<u><em>amount of moles of AgNO₃= 3.6 moles</em></u>

But 3.6 moles of AgNO₃ are not available, 2 moles are available. Since you have less moles than you need to react with 1.8 moles of Cu, AgNO₃ will be the limiting reagent.

<h3>Summary</h3>

In summary, AgNO₃ will be the limiting reagent.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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8 0
1 year ago
200.0 mL OF A 5.10 M solution are diluted to 500.0 mL. What is the concentration of the resulting solution? *
erastova [34]

Answer:

Concentration of the resulting solution = 2.04 M

Explanation:

<u>Data:</u>

M1 = 5.10 M

V1 = 200.0 mL

V2 = 500.0 mL

M2 = ?

By modifying the volume of solution, keeping the amount of solute constant, the concentration changes. To perform the calculations, the equation will be:

M1*V1=M2*V2

Where M1 is the initial concentration of the solution, M2 the final concentration and V is the value of the volumes of the initial and final solution.

Clearing the value of M2 from the equation and replacing the values ​​we have:

M2=\frac{M1*V1}{V2} =\frac{5.10M*200mL}{500mL} =2.04M

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Does anyone know this? help please
Sav [38]

Explanation:

The substance that is in excess that doesn't get used up as a reactant is called limiting reactant.

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