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3 years ago

Theoretical yield and percentage yield of:

Chemistry

1 answer:

dem82 [27]3 years ago

7 0

Answer:

Theoretical yield of potassium nitrate = 1.21 g

Theoretical yield of lead iodide = 2.77 g

Explanation:

Given data:

Mass of Lead nitrate = 2.00 g

Mass of potassium iodide = 3.00 g

Theoretical yield = ?

Percent yield = ?

Solution:

Chemical equation:

Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

Now we will calculate the number of moles of lead nitrate.

Number of moles = mass/ molar mass

Number of moles = 2.00 g/ 331.2 g/mol

Number of moles = 0.006 mol

Number of moles of KI:

Number of moles = mass/ molar mass

Number of moles = 3.00 g/ 166 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of lead nitrate with potassium iodide with lead iodide and potassium nitrate.

Pb(NO₃)₂ : PbI₂

1 : 1

0.006 : 0.006

Pb(NO₃)₂ : KNO₃

1 : 2

0.006 : 0.006×2 =0.012

KI : PbI₂

2 : 1

0.02 : 1/2×0.02 = 0.01

KI : KNO₃

2 : 2

0.02 : 0.02

Theoretical yield of both product depend upon lead nitrate because it is limiting reactant.

Theoretical yield of potassium nitrate:

Mass = number of moles × molar mass

Mass = 0.012 mol × 101.1 g/mol

Mass = 1.21 g

Theoretical yield of lead iodide:

Mass = number of moles × molar mass

Mass = 0.006 mol × 461.01 g/mol

Mass = 2.77 g

Percentage yield can not be determine because actual yield is not given.

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Describe the double circulation system of the human body

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At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

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Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

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A 0.0000792 M sample of Compound X in a solvent has an absorbance of 0.341 at 528 nm in a 1.000-cm cuvet. The solvent alone has

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Answer:

a) a = 3485 M⁻¹cm⁻¹

b) C = 0,000127 M

Explanation:

Lambert-Beer law says that there is a linear relationship between absorbance and concentration of a chemical substance. The formula is:

A = a×b×C

Where A is absorbance, a is molar absorptivity, b is path length and C is concentration.

a) In the problem Concentration is 0.0000792 M, b is 1,000cm and Absorbance is absorbance of sample-absorbance of blank: 0,341-0,065 = 0,276

Replacing:

0,276 = a×1,000cm×0,0000792M

a = 3485 M⁻¹cm⁻¹

b) As the experiment consist in the same compound in the same solvent, the molar absorptivity will be the same, a = 3485 M⁻¹cm⁻¹, path length will be 1,000cm and absorbance: 0,508-0,065 = 0,443

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3 years ago

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

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