Assuming 54.81 grams of Al are consumed in the presence of excess copper II chloride dihydrate, how many grams of AlCl3 can be produced if the reaction will only produce 66.93 % yield?

Answers

Answer 1

Answer:

181.39g of AlCl3 is produced

Explanation:

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

3CuCl2•2H2O + 2Al → 2AlCl3 + 6H2O + 3Cu

Next, we shall determine the mass of Al that reacted and the mass of AlCl3 produced from the balanced equation. This is illustrated below:

Molar mass of Al = 27g/mol

Mass of Al from the balanced equation = 2 x 27 = 54g

Molar mass of AlCl3 = 27 + (3x35.5) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Summary:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Next, we shall determine the theoretical yield of AlCl3. This can be achieved as shown below:

From the balanced equation above,

54g of Al reacted to produce 267g of AlCl3.

Therefore, 54.81g of Al will react to produce = (54.81 x 267)/54 = 271.01g of AlCl3.

Therefore, the theoretical yield of AlCl3 is 271.01g.

Finally, we shall determine the actual yield of AlCl3 produced from the reaction.

This can be obtain as follow:

Percentage yield of AlCl3 = 66.93%

Theoretical yield of AlCl3 = 271.01g

Actual yield of AlCl3 =?

Percentage yield = Actual yield/Theoretical yield x 100

66.93% = Actual yield /271.01g

Actual yield = 66.93% x 271.01

Actual yield = 66.93/100 x 271.01g

Actual yield = 181.39g.

Therefore, 181.39g of AlCl3 is produced from the reaction.

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How many moles of sulfur atoms are there in 5.0 g of sulfur?

Answers

Answer:

Number of moles = 0.153 mol

Explanation:

Given data:

Mass of sulfur = 5 g

Number of moles of sulfur atom = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of sulfur is 32. 065g/mol.

By putting values,

Number of moles = 5 g/ 32.06 g/mol

Number of moles = 0.153 mol

Estimate how much heat in joules is released when 25.0 g of water (C = 4.184 J/g°C) is cooled from 80.0°C to 30.0°C?

Answers

The amount of heat will be 5230 j.

What is heat?

Heat is a type of energy that is transferred between both the system and its surroundings as a result of temperature variations.

Calculation of heat.

Given data:

Mass = 25.0 g = 0.025 kg

C = 4.184 J/g°C

$T_(1)$ = 80.0°C

$T_(2)$ = 30.0°C

Q= ?

By using the formula of heat.

Q = MC ( $T_(2) - T_(1)$ )

Put the value of given data in heat equation.

Q(heat) = 0.025 × 4.184 ( 30 - 80)

Q(heat) = 5230 J.

Therefore, the amount of heat will be 5230 J.

To know more about heat.

brainly.com/question/13860901.

#SPJ2

Answer:

5230 J

Explanation:

m = 25 g = 0,025 kg

c = 4,184 J /(g * °C) = 4184 J /(kg * °C)

$t_(1)$ = 80 °C

$t_(2)$ = 30 °C

The formula is Q = c *m * ( $t_(2) - t_(1)$ )

Calculating:

Q = 4184 * 0,025 * (30 - 80) = 5230 J

Note that we get a negative heat (-5230 J). It just means that it is released.

What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing point depression constant for water is 1.86 °C/m.

Answers

Answer:

freezing point (°C) of the solution = - 3.34° C

Explanation:

From the given information:

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

where;

i = vant Hoff factor

the vant Hoff factor is the totality of the number of ions in the solution

Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3

$K_f$ = 1.86 °C/m

m = molality of the solution and it can be determined by using the formula

$molality = (mole \ of \ solute )/(kg \ of \ solvent )$

which can now be re-written as :

$molality = (mole \ of \ Ca(NO_3)_2)/(kg \ of \ water)$

$molality = ((mass \ of \ \ Ca(NO_3)_2)/(molar \ mass of \ Ca(NO_3)_2) )/(kg \ of \ water)$

$molality = ((11.3 \ g )/(164 \ g/mol) )/(0.115 \ kg )$

molality = 0.599 m

∴

The freezing point (°C) of a solution can be prepared by using the formula:

$\Delta T = iK_fm$

$\Delta T =3 * (1.86 \ ^0C/m) * (0.599 \ m)$

$\Delta T =3.34^0 \ C$

$\Delta T =$ the freezing point of water - freezing point of the solution

3.34° C = 0° C - freezing point of the solution

freezing point (°C) of the solution = 0° C - 3.34° C

freezing point (°C) of the solution = - 3.34° C

Salt, sugar, and snow are examples of ____________________ solids.

Answers

Crystalline I hope this helps:)

They are examples of Semi solids

I have always enjoyed eating tuna fish. Unfortunately,a study of the mercury content of canned tuna in 2010 foundthat chunk white tuna contains 0.6 ppm Hg and chunk lighttuna contains 0.14 ppm. (S. L. Gerstenberger, A. Martinson,and J. L. Kramer, Environ. Toxicol. Chem. 2010, 29, 237.) TheU.S. Environmental Protection Agency recommends no morethan 0.1 mg Hg/kg body weight per day. I weigh 68 kg. Howoften may I eat a can containing 6 ounces (1 lb 5 16 oz) ofchunk white tuna so that I do not average more than 0.1 mgHg/kg body weight per day? If I switch to chunk light tuna,how often may I eat one can?

Answers

Answer:

You can eat one chunk white tuna of 6 oz every 21,5 minutes

And, you can eat one chunk linght tuna of 6 oz every 5 minutes

Explanation:

The exposure to mercury may cause serious health problems, and is a threat to the development of the child in utero and early in life.

If you weight 68 kg you can eat:

68 kg * 0,1mgHg/ kg = 6,8 mg Hg per day

Chunk white tuna contains 0,6 mg Hg/kg

Thus, you can eat:

6,8 mg Hg * 1 kg tuna/ 0,6 mg Hg = 11,33 kg of chunk white tuna per day

In ounces:

11,33 kg * 35,274oz/ 1 kg = 400 oz per day

You can eat 66,7 6 ounces of chunk white tuna per day. One every 21,5 minutes

Chunk light tuna contains 0,14 mg Hg/kg

Thus, you can eat:

6,8 mg Hg * 1 kg tuna/ 0,14 mg Hg = 48,57 kg of chunk white tuna per day

In ounces:

48,57 kg * 35,274oz/ 1 kg = 1713 oz per day

You can eat 285,6 6 ounces of chunk white tuna per day. One every 5 minutes

I hope it helps!

What is the concentration of a solution that has 15.0 g NaCl dissolved to a total of 750 ml?

Answers

Answer: The concentration of solution is 0.342 M

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \text{Volume of solution (in mL)}}$

We are given:

Mass of solute (Sodium chloride) = 15 g

Molar mass of sodium chloride = 58.5 g/mol

Volume of solution = 750 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=(15g* 1000)/(58.5g/mol* 750mL)\n\n\text{Molarity of solution}=0.342M$

Hence, the concentration of solution is 0.342 M

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