Based on your lab report, provide the chemical reaction that occurs in Benedict's test for the following molecule.

For each of the reactions, calculate the mass (in grams) of the product formed when 15.12g of the underlined reactant completely

Ludmilka [50]

Answer:

Part A : amount of product (KCl) = 28.88 g

Part B : amount of product (KBr) = 46.13 g

Part C : amount of product (Cr₂O₃) = 17.3 g

Part D: amount of product (SrO) = 35.76 g

Explanation:

Part A:

Data Given:

Reaction :

2K(s) + Cl₂(g) --------> 2KCl

Amount of underline Reactant (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

2K(s) + Cl₂(g) --------> 2KCl

2mol 1mol 2mol

it shows that

2 mole of K give 2 mole of KCl

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KCl = (39 + 35.5)

Molar mass of KCl = 74.5 g/mol

So the look again to reaction in terms of grams

2K(s) + Cl₂(g) --------> 2KCl

2mole (39 g/mol) 2mole (74.5 g/mol)

78 g 149 g

Apply the Unity formula

78 g of Potassium ≅ 149 g of KCl

Then

15.12 g of Potassium ≅ how many g of Product (KCl)

By doing cross multiplication

X g of Product (KCl) = 149 g of KCl x 15.12 g of K / 78 g of K

X g of Product (KCl) = 28.88 g

So the amount of product (KCl) = 28.88 g

_________________________________________

Part B:

Data Given:

Reaction :

2K(s) + Br₂(g) --------> 2KBr

Amount of underline Reactant (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

2K(s) + Br₂(g) --------> 2KBr

2mol 1mol 2mol

it shows that

2 mole of K give 2 mole of KBr

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KBr = (39 + 80)

Molar mass of KBr = 119 g/mol

So, look again to reaction in terms of grams

2K(s) + Br₂(g) --------> 2KBr

2mole (39 g/mol) 2mole (119 g/mol)

78 g 238 g

Apply the Unity formula

78 g of Potassium ≅ 238 g of KBr

Then

15.12 g of Potassium ≅ how many g of Product (KBr)

By doing cross multiplication

X g of Product (KBr) = 238 g of KBr x 15.12 g of K / 78 g of K

X g of Product (KBr) = 46.13 g

So the amount of product (KBr) = 46.13 g

__________________________________________

Part C:

Data Given:

Reaction :

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

Amount of underline Reactant (Cr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.

So, amount of product depend on the amount of Chromium (Cr)

Now Look at Given Reaction:

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

4mol 3mol 2mol

it shows that

4 mole of Cr give 2 mole of Cr₂O₃

if we represent mole in grams

Then

Molar mass of Cr = 52 g/mol

Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)

Molar mass of 2Cr₂O₃ = 304 g/mol

So, look again to reaction in terms of grams

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

4 mol (52 g/mol) 2 mole (304 g/mol)

208 g 608 g

Apply the Unity formula

208 g of Chromium ≅ 608 g of Cr₂O₃

Then

15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)

By doing cross multiplication

X g of Product (Cr₂O₃) = 238 g of Cr₂O₃ x 15.12 g of Cr / 208 g of Cr

X g of Product (Cr₂O₃) = 17.3 g

So the amount of product (Cr₂O₃) = 17.3 g

________________________________________

Part D:

Data Given:

Reaction :

2Sr(s) + O₂(g) --------> 2SrO(s)

Amount of underline Reactant (Sr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.

So, amount of product depend on the amount of Strontium (Sr)

Now Look at Given Reaction:

2Sr(s) + O₂(g) --------> 2SrO(s)

2mol 1mol 2mol

it shows that

2 mole of Sr give 2 mole of SrO

if we represent mole in grams

Then

Molar mass of Sr = 87.6 g/mol

Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)

Molar mass of 2SrO = 207.2 g/mol

So, look again to reaction in terms of grams

2Sr(s) + O₂(g) --------> 2SrO(s)

2 mol ( 87.6 g/mol) 2 mole (207.2 g/mol)

175.2 g 414.4 g

Apply the Unity formula

175.2 g of Strontium ≅ 414.4 g of SrO

Then

15.12 g of Strontium ≅ how many g of Product (SrO)

By doing cross multiplication

X g of Product (SrO) = 414.4 g of SrO x 15.12 g of Sr / 175.2 g of Sr

X g of Product (SrO) = 35.76 g

So the amount of product (SrO) = 35.76 g

8 0

3 years ago

Predict the formula of the ionic compound that would result from combining the monatomic ions formed by each pair of elements. M

skad [1K]

Answer:

E - Be and O

A - Mg and N

E - Li and Br

F - Ba and Cl

B - Rb and O

Explanation:

Be and O

Be is a metal that loses 2 e⁻ to form Be²⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form BeO (E-MX).

Mg and N

Mg is a metal that loses 2 e⁻ to form Mg²⁺ and N is a nonmetal that gains 3 e⁻ to form O³⁻. For the ionic compound to be neutral, it must have the form Mg₃N₂ (A-M₃X₂).

Li and Br

Li is a metal that loses 1 e⁻ to form Li⁺ and Br is a nonmetal that gains 1 e⁻ to form Br⁻. For the ionic compound to be neutral, it must have the form LiBr (E-MX).

Ba and Cl

Ba is a metal that loses 2 e⁻ to form Ba²⁺ and Cl is a nonmetal that gains 1 e⁻ to form Cl⁻. For the ionic compound to be neutral, it must have the form BaCl₂ (F-MX₂).

Rb and O

Rb is a metal that loses 1 e⁻ to form Rb⁺ and O is a nonmetal that gains 2 e⁻ to form O²⁻. For the ionic compound to be neutral, it must have the form Rb₂O (B-M₂X).

4 0

3 years ago

Write the general electron configuration of a transition element(a) in Period 5;

atroni [7]

Electron configurations of transition elements in Period 5:

Y Yttrium [Kr] 4d1 5s2

Zr Zirconium [Kr] 4d2 5s2

Nb Niobium [Kr] 4d4 5s1 (*)

Mo Molybdenum [Kr] 4d5 5s1 (*)

Tc Technetium [Kr] 4d5 5s2

Ru Ruthenium [Kr] 4d7 5s1 (*)

Rh Rhodium [Kr] 4d8 5s1 (*)

Pd Palladium [Kr] 4d10 (*)

Ag Silver [Kr] 4d10 5s1 (*)

Cd Cadmium [Kr] 4d10 5s2

What are transition elements?

The transition elements or transition metals occupy the short columns in the center of the periodic table, between Group 2A and Group 3A.
They are sometimes called the d-block elements, since in this region the d-orbitals are being filled in, and are also referred to as B-group elements since in most numbering systems of the columns on the periodic table the numerals of these groups are followed by the letter B.

What are period 5 transition elements?

The period 5 transition metals are yttrium (Y), zirconium (Zr), niobium (Nb), molybdenum (Mo), technetium (Tc), ruthenium (Ru), rhodium (Rh), palladium (Pd), silver (Ag), and cadmium (Cd).

To know more about transition elements, refer:

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7 0

2 years ago

Which statements accurately describe soap? Select one or more:

tia_tia [17]

Answer:

A. Soaps react with ions in hard water to create a precipitate.

B. Soaps are both hydrophobic and hydrophilic.

D. Soaps should be weakly alkaline in solution.

Explanation:

A. Hard water contains <u>magnesium and calcium minerals</u> like calcium and magnesium carbonates, sulfates and bicarbonates. As soon as these minerals come in contact with soap their ions like Mg²⁺ & Ca²⁺ form precipitates.

B. Soap are both hydrophilic and hydrophobic. They reason why they exhibit both the properties is really important for their functionality. The hydrophobic part of soap makes interaction with oil/dust particles while the hydrophilic part makes interaction with water. When the cloth is rinsed the dirt/soap particles are removed from the dirty clothes thereby making them clean.

C. Soaps have alkaline pH i.e. more than 7 that is why they have bitter taste.

5 0

3 years ago

How many moles of solute particles are present in 1 mL (exact) of aqueous 0.0040 M Ba(OH)2?

lidiya [134]

The number of mole of solute particles are present in 1 mL (exact) of aqueous 0.0040 M Ba(OH)₂ is 0.000004 mole

<h3>What is molarity? </h3>

Molarity is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:

Molarity = mole / Volume

With the above formula, we can determine the number of mole present in the solution. Detail below:

<h3>How to determine the mole of the solute in the solution</h3>

Volume of solution = 1 mL = 1 / 1000 = 0.001 L
Molarity of solution = 0.004 M
Mole of solute =?

Molarity = mole / Volume

Cross multiply

Mole = Molarity × volume

Mole of solute = 0.004 × 0.001

Mole of solute = 0.000004 mole

Thus, the mole of the solute in the solution is 0.000004 mole

Learn more about molarity:

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6 0

2 years ago

Based on your lab report, provide the chemical reaction that occurs in Benedict's test for the following molecule.​

Based on your lab report, provide the chemical reaction that occurs in Benedict's test for the following molecule.