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11Alexandr11 [23.1K]
3 years ago
5

What can you predict about an element from its position in the periodic table?

Chemistry
1 answer:
Igoryamba3 years ago
8 0
Yeah of-course!! It's valency by group most of the chemical property like electronegativity, ionization energy etc. by the combination of groups and periods...
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Which of the following would be considered a strong base?
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Ca(OH)2 is a strong base.

It’s name is calcium hydroxide and often it is easy to tell if something is a strong base if it has (OH) in the name.
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URGENT!!! A car engine uses the combustion of fuel to move a system of gears that move its wheels. Not all of the fuel's energy
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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 $\large \boxed{\mathbf{98.67 \, \%}}$}

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