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faltersainse [42]
1 year ago
5

What is the percent yield of this reaction if 22.o g of Mgl2 is produced by the reaction of 25.0 g of Mg with 25.0 g of l2?

Chemistry
1 answer:
expeople1 [14]1 year ago
8 0

% yield = 80.719

<h3>Further explanation</h3>

Given

22.0 g of Mgl₂

25.0 g of Mg

25.0 g of l₂

Required

The percent yield

Solution

Reaction

Mg + I₂⇒ MgI₂

mol Mg = 25 g : 24.305 g/mol = 1.029

mol I₂ = 25 g : 253.809 g/mol = 0.098

Limiting reactant = I₂

Excess reactant = Mg

mol MgI₂ based on I₂, so mol MgI₂ = 0.098

Mass MgI₂ (theoretical):

= mol x MW

= 0.098 x 278.114

= 27.255 g

% yield = (actual/theoretical) x 100%

% yield = (22 / 27.255) x 100%

% yield = 80.719

You might be interested in
A 1.44-g sample of an unknown pure gas occupies a volume of 0.335 L at a pressure of and a temperature of 100.0°C. The unknown g
Zepler [3.9K]

Answer:

Xenon

Explanation:

Step 1: Given data

  • Mass (m): 1.44 g
  • Volume (V): 0.335 L
  • Pressure (P): 1.00 atm (I looked it up)
  • Temperature (T): 100.0°C

Step 2: Convert the temperature to Kelvin

K = °C + 273.15 = 100.0°C + 273.15 = 373.2 K

Step 3: Calculate the number of moles (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.00 atm × 0.335 L / (0.0821 atm.L/mol.K) × 373.2 K

n = 0.0109 mol

Step 4: Calculate the molar mass of the gas

M = 1.44 g / 0.0109 mol = 132 g/mol

Step 5: Identify the gas

The gas with a molar mass of about 132 g/mol is xenon.

8 0
2 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

Volume of the sample = V = 15.3 cm^3

Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

3 0
2 years ago
Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No
liraira [26]

Answer:

  • Molality = 0.066 m
  • Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The <em>freezing-point depression</em> is expressed by:

  • ΔT=Kf * m

We put the data given by the problem and <u>solve for m</u>:

  • 2.63 °C = 40°C·kg/mol * m
  • m = 0.06575 m

For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:

  • 0.06575 m = Moles reserpine / kg camphor
  • 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We<u> calculate moles of reserpine:</u>

  • 0.06575 m = Moles reserpine / 0.025 kg camphor
  • Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:

  • 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>

8 0
2 years ago
Use enthalpies of formation given in appendix c to calculate δh for the reaction br2(g)→2br(g), and use this value to estimate t
Contact [7]

Given reaction represents dissociation of bromine gas to form bromine atoms

Br2(g) ↔ 2Br(g)

The enthalpy of the above reaction is given as:

ΔH = ∑n(products)ΔH^{0}f(products) - ∑n(reactants)ΔH^{0}f(reactants)

where n = number of moles

ΔH^{0}f= enthalpy of formation

ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol

Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol

3 0
2 years ago
A penny has a mass of
Ahat [919]
You did not include the questions.

I did some research and found the questions:

<span> What is the mass of 1 mole of pennies? How many moles of pennies have a mass equal to the mass of the moon?

Solutions:

1) mass of 1 mole of pennies

Data: mass of 1 penny = 2.50 g

1 mole = 6.022 * 10^ 23 units

Proportion:

  1 penny      6.022 * 10^23 penny
-------------- = ----------------------------
   2.50 g                    x

Solve: x = 6.022 * 10^23 penny * 2.50g / 1 penny = 15.055* 10^23

Since 2.50 has 3 significant figures, the answer must use 3 significant figures => x = 15.1 * 10^ 23 g = 1.51 * 10^24 g

Answer: 1 mol of pennies have a mass of 1.51 * 10^24 g

2) How many moles of pennies have a mass equal to the same mass of the Moon

Convert the mass of the Moon grams: 7.35 * 10^22 kg = 7.35 * 10^ 25 g

       1 mol                            x
---------------------- =  ----------------------
1.51 * 10^ 24g          7.35 * 10^ 25 g

=> x = 7.35 * 10^ 25 g * 1 mol / (1.51 * 10^24 g)= 48.7 mol

Answer: 48.7 mol
</span>
5 0
2 years ago
Read 2 more answers
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