A salt solution with a volume of 250 mL that contains 0.70 mol NaCl.

Following the instructions in your lab manual, you have titrated a 25.00 mL sample of 0.0100 M KIO3 with a solution of Na2S2O3 o

vivado [14]

Answer:

For 1: The amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2: The amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

Explanation:

For 1:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$

Molarity of $KIO_3$ solution = 0.0100 M

Volume of solution = 25 mL

Putting values in above equation, we get:

$0.0100M=\frac{\text{Moles of }KIO_3\times 1000}{25}\\\\\text{Moles of }KIO_3=\frac{0.0100\times 25}{1000}=0.00025mol$

Hence, the amount of potassium iodate that were titrated is $2.5\times 10^{-4}$ moles

For 2:

The chemical equation for the reaction of potassium iodate and sodium thiosulfate follows:

$2KIO_3+Na_2S_2O_3\rightarrow K_2S_2O_3+2NaIO_3$

By Stoichiometry of the reaction:

2 moles of potassium iodate reacts with 1 mole of sodium thiosulfate

So, 0.00025 moles of potassium iodate will react with = $\frac{1}{2}\times 0.00025=0.000125mol$ of sodium thiosulfate

Hence, the amount of sodium thiosulfate required is $1.25\times 10^{-4}$ moles

6 0

3 years ago

A propane stove burned 470 grams propane and produced 625 grams of water (this is the actual yield) C3H8 +5O2=3CO2+4H20. What wa

Liula [17]

Answer:

81.3%

Explanation:

Step 1:

The balanced equation for the reaction:

This is shown below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Step 2:

Data obtained from the question. This includes:

Mass of propane (C3H8) = 470 g

Actual yield of water (H2O) = 625 g

Percentage yield of water (H2O) =?

Step 3:

Determination of the mass of propane (C3H8) burned and the mass of water (H2O) produce from the balanced equation. This is illustrated below:

C3H8 + 5O2 —> 3CO2 + 4H2O

Molar Mass of C3H8 = (3x12) + (8x1) = 36 + 8 = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 4 x 18 = 72g

From the balanced equation above,

44g of C3H8 was burned and 72g of H2O was produced.

Step 4:

Determination of the theoretical yield of H2O. This is illustrated below:

From the balanced equation above,

44g of C3H8 produced 72g of H2O.

Therefore, 470g of C3H8 will produce = (470x72)/44 = 769.09g of H2O.

Therefore, the theoretical yield of H2O is 769.09g

Step 5:

Determination of the percentage yield of water (H2O). This is illustrated below:

Actual yield of water (H2O) = 625g

theoretical yield of H2O = 769.09g

Percentage yield of water (H2O) =?

Percentage yield = Actual yield/Theoretical yield x100

Percentage yield = 625/769.09 x100

Percentage yield = 81.3%

Therefore, the percentage yield of water (H2O) is 81.3%

4 0

3 years ago

Find the percent composition of each element in KMnO4

VARVARA [1.3K]

The way you want to find the percent composition would be by breaking down the problem like so:

K= atomic mass of K which is 39.098
Mn = atomic mass of Mn which is 54.938
O= atomic mass of o which is 15.999

Then you want to add 39.098+ 54.938+ 15.999 and you get 110.035 which is the molar mass for KMnO

Then you want to take each molar mass and then divide it 110.035 and multiply by 100

Ex. K = 39.098/ 110.035 and the multiply what you get by a 100

You do this for the other elements as well good luck!

6 0

3 years ago

Near the surface of a liquid, fast-moving particles can break free and become a gas.

Nookie1986 [14]

This statement is True! Lets think about it... When water boils, the water doesnt evaporate from the bottom of the bowl it evaporates from the top!
=)

6 0

3 years ago

Determine the number of grams in each of the quantities 1.39.0 x 1024 molecules Cl2

STatiana [176]

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions) in a substance

This refers to the atomic total of the 12 gr C-12 which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

$\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31$

mass Cl₂(MW=71 g/mol) :

$\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01$

8 0

3 years ago