
An element E reacts with oxygen to form an oxide $ {{E}_{2}}O $ . What is its relative atomic mass if 6.9 g of the element reacts with water to form $ 1800\text{ }c{{m}^{3}} $ of hydrogen?
Answer
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Hint: A chemical reaction is a process that results in the chemical change of one set of chemical substances into another set of chemical substances. Chemical reactions are often defined as changes in the locations of electrons in the formation and breaking of chemical bonds between atoms, with no change in the nuclei (no change in the elements present), and may be represented using a chemical equation.
Complete answer:
At normal temperature and pressure, the alkali metals are all lustrous, soft, and extremely reactive metals that quickly shed their outermost electrons to create cations with charge +1. Because of their suppleness, they can all be sliced easily with a knife, revealing a gleaming surface that tarnishes quickly in the air owing to oxidation by atmospheric moisture and oxygen (and in the case of lithium, nitrogen). They must be kept under oil to avoid interaction with air due to their extreme reactivity, and they are only found in nature as salts, never as free elements.
$ 4E\text{ }+\text{ }2{{O}_{2}}\to 2{{E}_{2}}O $
Because it takes two atoms of E to mix with one atom of O, E is a Group 1 element.
Its hydroxide must have the formula EOH.
Then there's the water reaction $ 2E\text{ }+\text{ }2{{H}_{2}}O\to 2EOH+{{H}_{2}} $
The mole is the International System of Units' basic unit of material quantity (SI). It is defined as a collection of precisely $ 6.02214076\times {{10}^{23}} $ particles, which may be atoms, molecules, ions, or electrons. The Avogadro number ( $ 6.02214076\times {{10}^{23}} $ ) was set such that the mass of one mole of a chemical compound in gram is numerically equivalent to the average mass of one molecule of the compound in daltons for most practical applications.
$ \text{Number of moles = Given volume }\!\!\times\!\!\text{ }\dfrac{\text{Given mole}}{\text{Molar volume}} $
Moles of $ \mathrm{H}_{2}=1800 \mathrm{~cm}^{3} \mathrm{H}_{2} \times \dfrac{1 \mathrm{~mol} \mathrm{H}_{2}}{22400 \mathrm{~cm}^{3} \mathrm{H}_{2}}=0.0804 \mathrm{~mol} \mathrm{H}_{2} $
Moles of $ \mathrm{E}=0.0804 \mathrm{~mol} \mathrm{H}_{2} \times \dfrac{2 \mathrm{~mol} \mathrm{E}}{1 \mathrm{~mol} \mathrm{H}_{2}}=0.161 \mathrm{~mol} \mathrm{E} $
Molar mass of $ \mathrm{E}=\dfrac{6.9 \mathrm{~g}}{0.161 \mathrm{~mol}}=43 \mathrm{~g} / \mathrm{mol} $
$ A_{\mathrm{r}}=43 $
Hence atomic mass of E is 43.
Note:
Because of their extreme reactivity, alkali metals do not exist in pure form in nature. They are lithophiles, meaning they mix easily with oxygen and hence firmly interact with silica, creating relatively low-density minerals that do not sink deep into the Earth's core. Due to their high ionic radii, potassium, rubidium, and caesium are also incompatible elements.
Complete answer:
At normal temperature and pressure, the alkali metals are all lustrous, soft, and extremely reactive metals that quickly shed their outermost electrons to create cations with charge +1. Because of their suppleness, they can all be sliced easily with a knife, revealing a gleaming surface that tarnishes quickly in the air owing to oxidation by atmospheric moisture and oxygen (and in the case of lithium, nitrogen). They must be kept under oil to avoid interaction with air due to their extreme reactivity, and they are only found in nature as salts, never as free elements.
$ 4E\text{ }+\text{ }2{{O}_{2}}\to 2{{E}_{2}}O $
Because it takes two atoms of E to mix with one atom of O, E is a Group 1 element.
Its hydroxide must have the formula EOH.
Then there's the water reaction $ 2E\text{ }+\text{ }2{{H}_{2}}O\to 2EOH+{{H}_{2}} $
The mole is the International System of Units' basic unit of material quantity (SI). It is defined as a collection of precisely $ 6.02214076\times {{10}^{23}} $ particles, which may be atoms, molecules, ions, or electrons. The Avogadro number ( $ 6.02214076\times {{10}^{23}} $ ) was set such that the mass of one mole of a chemical compound in gram is numerically equivalent to the average mass of one molecule of the compound in daltons for most practical applications.
$ \text{Number of moles = Given volume }\!\!\times\!\!\text{ }\dfrac{\text{Given mole}}{\text{Molar volume}} $
Moles of $ \mathrm{H}_{2}=1800 \mathrm{~cm}^{3} \mathrm{H}_{2} \times \dfrac{1 \mathrm{~mol} \mathrm{H}_{2}}{22400 \mathrm{~cm}^{3} \mathrm{H}_{2}}=0.0804 \mathrm{~mol} \mathrm{H}_{2} $
Moles of $ \mathrm{E}=0.0804 \mathrm{~mol} \mathrm{H}_{2} \times \dfrac{2 \mathrm{~mol} \mathrm{E}}{1 \mathrm{~mol} \mathrm{H}_{2}}=0.161 \mathrm{~mol} \mathrm{E} $
Molar mass of $ \mathrm{E}=\dfrac{6.9 \mathrm{~g}}{0.161 \mathrm{~mol}}=43 \mathrm{~g} / \mathrm{mol} $
$ A_{\mathrm{r}}=43 $
Hence atomic mass of E is 43.
Note:
Because of their extreme reactivity, alkali metals do not exist in pure form in nature. They are lithophiles, meaning they mix easily with oxygen and hence firmly interact with silica, creating relatively low-density minerals that do not sink deep into the Earth's core. Due to their high ionic radii, potassium, rubidium, and caesium are also incompatible elements.
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