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3 years ago

When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction:

Chemistry

2 answers:

slamgirl [31]3 years ago

8 0

Answer:

1.5 mole

Explanation:

Step 1:

The balanced equation for the reaction. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

Step 2:

Determination of the masses of Al and Cl2 that reacted 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 Cl2 = 2 x 35.5 = 71g/mol

Mass of Cl2 from the balanced equation = 3 x 71 = 213g

From the balanced equation,

54g of Al reacted.

213g of Cl2 reacted

Step 3:

Determination of the limiting reactant.

This is illustrated below:

From the balanced equation above,

54g of Al reacted with 213g of Cl2.

Therefore, 40.5g of Al will react with = (40.5 x 213)/54 = 159.75g of Cl2.

From the calculations made above, there are leftover of Cl2 as 159.75g reacted out of 212.7g. Therefore, Cl2 is the excess reactant and Al is the limiting reactant.

Step 4:

Determination of the number of mole in 40.5g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 40.5g

Number of mole of Al =?

Number of mole = Mass/Molar Mass

Number of mole of Al = 40.5/27

Number of mole of Al = 1.5 mole

Step 5:

Determination of the number of mole of AlCl3 produced When 40.5 g of Al and 212.7 g of Cl2 combine together. This is illustrated below:

2Al(s) + 3Cl2(g) --> 2AlCl3(s)

From the balanced equation above,

2 moles of Al produced 2 moles of AlCl3.

Therefore, 1.5 mole of Al will also produce 1.5 mole of AlCl3.

From the calculations made above, 1.5 mole of AlCl3 is produced When 40.5 g of Al and 212.7 g of Cl2 combine together.

KIM [24]3 years ago

8 0

Answer:

1.5 moles AlCl3 will be formed

Explanation:

Step 1: Data given

MAss of Al = 40.5 grams

Mass of Cl2 = 212.7 grams

Atomic mass Al = 26.98 g/mol

Molar mass Cl2 = 70.9 g/mol

Step 2: The balanced equation

2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s)

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al = 40.5 grams / 26.98 g/mol

Moles Al = 1.50 moles

Moles Cl2 = 212.7 grams / 70.9 g/mol

Moles Cl2 = 3.0 moles

Step 4: Calculate limiting reactant

For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

Al is the limiting reactant. It will completely be consumed (1.5 moles).

Cl2 is in excess. There will react 3/2 * 1.5 = 2.25 moles Cl2

There will remain 3.0 - 2.25 = 0.75 moles Cl2

Step 5: Calculate moles AlCl3

For 2 moles Al we need 3 moles Cl2 to produce 2 moles AlCl3

For 1.5 moles AlC we'll have 1.5 moles AlCl3

1.5 moles AlCl3 will be formed

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4 years ago

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What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigo- nal planar arra

Reika [66]

Answer:

sp³;

sp²;

sp;

None;

One;

Two;

They're used to pi bonds.

Explanation:

The central atom in a molecule is generally the one that can make a greater number of bonds. The covalent bonds are made by the sharing of electrons, and, for that, the electron must be alone in the orbital.

To explain this, the hybridization theory was created, which states that, the orbitals are joined to form hybrids ones, and so, by the Hund's law, the electrons are alone in them.

The sigma bonds are done in the hybrids orbitals, and at the pure orbitals, the pi bonds are done. The lone pair of electrons are at a pure orbital. So, to know the hybridization of the central atom, we must know how many sigma bonds it does, and it will be the number of hybrids orbitals (each orbital may have two electrons, thus each bond are done in one orbital).

Double bonds and triple bonds have always only one sigma bond, so the number of sigma bonds is equal to the number of bonds, it's not necessary to know if they are simple, double or triple.

When the arrangement is tetrahedral, the central atom does 4 bonds, so it has 4 sigma bonds, and 4 hybrids orbitals (one of s and three for p), does its hybridization is sp³. Because exists only 3 p orbitals, there are no unhybridized p orbitals in this case.

When the arrangement is trigonal, the central atom does 3 bonds, so it has 3 hybrids orbitals (one of s and two of p), thus the hybridization is sp². So there are one unhybridized p orbitals.

When the arrangement is linear, the central atom does 2 bonds, so it has 2 hybrids orbitals (one of s and one of p), thus the hybridization is sp. So, there are two unhybridized p atoms.

As stated before, the unhybridized p orbitals are used to pi bonds.

5 0

3 years ago

Please help , science tho. by the way i have to do 5

kap26 [50]

Answer:

850

Explanation:

It's between 840 and 860

8 0

2 years ago

Mr. Chem S. Tree added water to 250. ML of a 2.50 M NaOH solution, until the final volume was 500. ML. What is the new molarity

tresset_1 [31]

Answer:

molarity of diluted solution = 1.25 M

Explanation:

Using,

C1V1 (Stock solution) = C2V2 (dilute solution)

given that

C1 = 2.50M

V1 = 250ML

C2 = ?

V2 = 500ML

2.50 M x 250 mL = C2 x 500 mL

C2 = (2.50 M x 250 mL) / 500 mL

C2 = 1.25 M

Hence, molarity of diluted solution = 1.25 M

4 0

3 years ago

Thorium-234 has a half-life of 24.1 days. How many grams of a 150 g sample would you have after 120.5 days?

chubhunter [2.5K]

Answer:

8.625 grams of a 150 g sample of Thorium-234 would be left after 120.5 days

Explanation:

The nuclear half life represents the time taken for the initial amount of sample to reduce into half of its mass.

We have given that the half life of thorium-234 is 24.1 days. Then it takes 24.1 days for a Thorium-234 sample to reduced to half of its initial amount.

Initial amount of Thorium-234 available as per the question is 150 grams

So now we start with 150 grams of Thorium-234

$150 \times \frac{1}{2}=24.1$