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Strike441 [17]
4 years ago
7

The ionization energies in kcal for mg and ca are mg ca 1st 176 141 2nd 346 247 3rd 1847 a) indicate the reasons for the relativ

ely high 3rd ionization energy for magnesium. b) indicate the reasons for the differences in 1st ionization energies for mg and ca.
Chemistry
1 answer:
Klio2033 [76]4 years ago
3 0

a) Answer is: the reasons for the relatively high 3rd ionization energy for magnesium is necause magnesium cation Mg²⁺ has stable electron configuration as noble gas neon and hardly lose electron.

Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.

b) Answer is: calcium has lower 1st ionization energy because valence electron are farther from nucleaus and interaction between protons and electrons are weaker.

Atomic number of magnesium (Mg) is 12, it means that it has 12 protons and 12 electrons.  

Electron configuration of magnesium atom: ₁₂Mg 1s² 2s² 2p⁶ 3s².  

Atomic number of calcium is 20, it means that it has 20 protons and 20 electrons, so atom of calcium is neutral.  

Electron configuration of calcium atom: ₂₀Ca 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².

Magnesium (Mg) and calcium (Ca) are metals from 2. group of Periodic table of elements and have low ionisation energy and electronegativity, which means they easily lose valence electons (two valence electrons).

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

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We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

MM:        17.03    32.00     30.01

              4NH₃  +  5O₂ ⟶ 4NO + 6H₂O

Mass/g:    1.5        1.85

2. Calculate the moles of each reactant  

\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}

3. Calculate the moles of NO we can obtain from each reactant

From NH₃:

The molar ratio is 4 mol NO:4 mol NH₃

\text{Moles of NO} = \text{0.0881 mol NH}_{3} \times \dfrac{\text{4 mol NO}}{\text{4 mol NH}_{3}} = \text{0.0881 mol NO}

From O₂:

The molar ratio is 4 mol NO:5 mol O₂

\text{Moles of NO} =  \text{0.057 81 mol O}_{2}\times \dfrac{\text{4 mol NO}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NO}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO.

The excess reactant is NH₃.

5. Calculate the mass of NO formed

\text{Mass of NO} = \text{0.046 25 mol NO}\times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \textbf{1.39 g NO}

6. Calculate the moles of NH₃ reacted

The molar ratio is 4 mol NH₃:5 mol O₂

\text{Moles reacted} = \text{0.057 81 mol O}_{2} \times \dfrac{\text{4 mol NH}_{3}}{\text{5 mol O}_{2}} = \text{0.046 25 mol NH}_{3}

7. Calculate the mass of NH₃ reacted

\text{Mass reacted} = \text{0.046 25 mol NH}_{3} \times \dfrac{\text{17.03 g NH}_{3}}{\text{1 mol NH}_{3}} = \text{0.7876 g NH}_{3}

8. Calculate the mass of NH₃ remaining

Mass remaining = original mass – mass reacted = (1.5 - 0.7876) g = 0.7 g NH₃

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T = temperature of eh reaction

ΔS° = Entropy change

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