All categories

2 years ago

Which formula can be used to calculate the theoretical yield?

Chemistry

1 answer:

Bas_tet [7]2 years ago

5 0

Answer:

Option C, (Actual yield ÷ percent yield) × 100

Explanation:

Theoretical yield is defined as the total amount of product formed for given reactants in a chemical reaction. It is an ideal case which assumes no exceptions or wastage.

The mathematical relation between the actual yield, percent yield and theoretical yield is as follows -

$P.Y. = \frac{M_{A.Y.}}{M_{T.Y.}} * 100$

Where

P.Y. represents the percent yield a

M A.Y. represents the mass obtained from actual yield

M T.Y. represents the mass obtained from theoretical yield

Hence, if we rearrange the formula, we get -

$M_{T.Y.} = \frac{M_{A.Y.}}{P.Y.} * 100$

Hence, option C is correct

You might be interested in

What is a function of chloroplast

Viktor [21]

The primary function of chloroplast is to conduct photosynthesis, which is the process plants use to convert light energy into chemical energy that can be released to fuel the organisms' activities.

8 0

3 years ago

Read 2 more answers

Fritz-Haber process

maks197457 [2]

Answer:

5×10⁵ L of ammonia (NH3)

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above, we can say that:

3 L of H2 reacted to produce 2 L of NH3.

Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:

From the balanced equation above,

3 L of H2 reacted to produce 2 L of NH3.

Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.

Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.

6 0

2 years ago

Which molecule has a dipole moment of 0 d?

fiasKO [112]

The answer is

a symmetrical molecule.

5 0

2 years ago

Which of the following statements BEST explains dissolution?

nordsb [41]

I had the same question, it's most likely B.

5 0

2 years ago

If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of

Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.1\ g}{186.04\ g/mol}$

$Moles\ of\ ferrocene= 0.0005375\ mol$

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

$\rho=\frac{Mass}{Volume}$

or,

$Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g$

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.044\ g}{78.49\ g/mol}$

$Moles\ of\ acetyl\ chloride= 0.0005606\ mol$

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene

Moles of product formed = 0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

<u>% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %</u>

5 0

3 years ago