Bonds payable issued with collateral are called ______

Chromium is dissolved in sulfuric acid according to the following equation: Cr + H2SO4 ⇒ Cr2 (SO4) 3 + H2

Usimov [2.4K]

Answer:

$\large \boxed{\text{a)188.4 g; b) 98.67 $\, \%$}}$

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 98.08 392.18

2Cr + 3H₂SO₄ ⟶ Cr₂(SO₄)₃ + 3H₂

To solve the stoichiometry problem, you must

Use the molar mass of H₂SO₄ to convert the mass of H₂SO₄ to moles of H₂SO₄
Use the molar ratio to convert moles of H₂SO₄ to moles of Cr₂(SO₄)₃
Use the molar mass of Cr₂(SO₄)₃ to convert moles of Cr₂(SO₄)₃ to mass of Cr₂(SO₄)₃

a) Mass of Cr₂(SO₄)₃

(i) Mass of pure H₂SO₄

$\text{Mass of pure} = \text{165 g impure} \times \dfrac{\text{85.67 g pure} }{\text{100 g impure}} = \text{141.36 g pure}$

(ii) Moles of H₂SO₄

$\text{Moles of H$_{2}$SO}_{4} = \text{141.36 g H$_{2}$SO}_{4} \times \dfrac{\text{1 mol H$_{2}$SO}_{4}}{\text{98.08 g H$_{2}$SO}_{4}} = \text{1.441 mol H$_{2}$SO}_{4}$

(iii) Moles of Cr₂(SO₄)₃

The molar ratio is 1 mol Cr₂(SO₄)₃:3 mol H₂SO₄ $\text{Moles of Cr$_{2}$(SO$_{4}$)}_{3} = \text{1.441 mol H$_{2}$SO}_{4} \times \dfrac{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}}{\text{3 mol H$_{2}$SO}_{4}} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3}$

(iv) Mass of Cr₂(SO₄)₃ $\text{Mass of Cr$_{2}$(SO$_{4}$)}_{3} = \text{0.4804 mol Cr$_{2}$(SO$_{4}$)}_{3} \times \dfrac{\text{392.18 g Cr$_{2}$(SO$_{4}$)}_{3}}{\text{1 mol Cr$_{2}$(SO$_{4}$)}_{3}} = \textbf{188.4 g Cr$_{2}$(SO$_{4}$)}_{3}\\\text{The mass of Cr$_{2}$(SO$_{4}$)$_{3}$ formed is $\large \boxed{\textbf{188.4 g}}$}$

b) Percentage yield

It is impossible to get a yield of 485.9 g. I will assume you meant 185.9 g.

$\text{Percentage yield} = \dfrac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \, \% = \dfrac{\text{185.9 g}}{\text{188.4 g}} \times 100 \, \% = \mathbf{98.67 \, \%}\\\\\text{The percentage yield is $\large \boxed{\mathbf{98.67 \, \%}}$}$

7 0

4 years ago

How many Hydrogen atoms are there in 5 g of NH3? (molar mass of NH3 = 17<br>g/mol)

goldfiish [28.3K]

Answer:5.309 × 10²⁴ atoms.

Explanation:

Given that

molar mass of NH3 = 17 g/mol

Mass of NH3 = 5g

Therefore, No of moles of NH3 = Mass/ molar mass

= 5g/ 17g/mol

= 0.294 moles.

I mole = 6.02 × 10²³ atoms

Therefore the number of hydrogen atoms in a 0.294 moles of ammonia gives us

0.294× 6.02 × 10²³ × 3 ( since there are 3 hydrogens in Ammonia )

= 5.309 × 10²⁴ atoms.

5 0

3 years ago

Heyyyyyyyyyyyyyyyyyyyyyyyyyyy

mart [117]

Answer:

hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Explanation:

7 0

3 years ago

Read 2 more answers

What are the isomers of c3h6br2...

maks197457 [2]

C3H6Br2 2 isomers:- bominated allkane

CH3CH2CHBr2 1,1-dibromopropane
<span>CH3CHBrCH2Br 1,2-dibromopropane

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>

7 0

4 years ago

The process by which nitrogen gas is converted to a usable form is called nitrogen

LenaWriter [7]

Answer: i beleive it is fixation in edge 2020

Explanation:

4 0

3 years ago

Read 2 more answers

Bonds payable issued with collateral are called _______ bonds.