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postnew [5]
4 years ago
12

How do alkali metals differ from alkaline earth metals in atomic structure and general properties?

Chemistry
1 answer:
Temka [501]4 years ago
6 0

Answer:

In the modern period table, the elements are arranged in the increasing order and depending on their physical and chemical properties they are classified into the different groups and atoms.

Alkali metals differ from alkaline earth metal in the atomic structure and their properties as all the alkali metals consist electron in the outer-most shell. And the earth alkaline metals has two outer electron in their shell. So, for achieving the configuration of noble gas the alkali metals loses one electron and the alkaline metal lose two electrons.  

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The number of electrons in a neutral atom is equal to the number of protons. The mass number of the atom (M) is equal to the sum of the number of protons and neutrons in the nucleus. The number of neutrons is equal to the difference between the mass number of the atom (M) and the atomic number (Z).

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Answer :

Part 1 : Balanced reaction, 3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

Part 2 : The theoretical yield of NH_3 gas = 440.96 g

Part 3 : The % yield of ammonia is 90.03 %

Solution : Given,

Mass of N_2 = 475 g

Molar mass of N_2 = 28 g/mole

Molar mass of NH_3 = 17 g/mole

Experimental yield of NH_3 = 397 g

<u>Answer for Part (1) :</u>

The balanced chemical reaction is,

3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

<u>Answer for Part (2) :</u>

First we have to calculate the moles of N_2.

\text{Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molecular mass of }N_2}=\frac{475g}{28g/mole}=16.96moles

From the given reaction, we conclude that

1 moles of N_2 gas react to give 2 moles of NH_3 gas

16.96 moles of N_2 gas react to give \frac{2}{1}\times 16.96=33.92 moles of NH_3 gas

Now we have to calculate the mass of NH_3 gas.

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(33.92moles)\times (17g/mole)=440.96g

Therefore, the theoretical yield of NH_3 gas = 440.96 g

<u>Answer for Part (3) :</u>

Formula used for percent yield :

\% \text{ yield of }NH_3=\frac{\text{ Experimental yield of }NH_3}{\text{ Theoretical yield of }NH_3}\times 100

\% \text{ yield of }NH_3=\frac{397g}{440.96g}\times 100=90.03\%

Therefore, the % yield of ammonia is 90.03 %

3 0
3 years ago
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