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What is the mass of \[2.6\]moles of\[{\text{Ag}}\]?

Answer
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Hint: The mole is one of the important units in chemistry. All the chemical reactions are expressed in terms of the moles of the reactant and product. If one reaction is completed or not also determined by the moles of the reactant and product. In the periodic table, each and every element has a unique symbol. In that case, \[{\text{Ag}}\] is the symbol of a silver atom.

Formula used: We have to know that the moles is defined as the given mass of the atom divided by the atomic weight of the atom.
\[{\text{Moles = }}\dfrac{{{\text{given weight of the atom}}}}{{{\text{atomic weight of the atom}}}}\]
\[{\text{Given Weight Of The Atom = number of moles}} \times {\text{atomic weight of the atom}}\]
If we want the given weight of the atom means the product of the number of moles of the atom and the atomic weight of the atom.

Complete step by step answer:
Given data:
The number of moles of silver is \[107.868g\].
The atomic weight of the silver is \[2.6\].
Calculate the mass of \[2.6\]moles of \[{\text{Ag}}\] is
\[{\text{Moles = }}\dfrac{{{\text{given weight of the atom}}}}{{{\text{atomic weight of the atom}}}}\]
We alter the formula depend on our need,
\[{\text{Given Weight Of The Atom = number of moles}} \times {\text{atomic weight of the atom}}\]
Now we can substitute the given values we get,
\[ = 2.6 \times 107.868\]
On multiplication we get,
\[ = 280.4568\]
From above calculation,
The mass of \[2.6\] moles of \[{\text{Ag}}\] is \[280.4568g\]

Note: As we know that the atomic number of silver is \[47\]. The atomic number of the atom is equal to the number of the electrons or number of protons in the atom. Therefore, \[47\] number of electrons and \[47\] number of protons are present in the silver atom. Because, the atomic number of silver is \[47\]. The mass number of the atom is nothing but the number of protons and number of electrons in the atom. Silver is one of the transition elements in the periodic table. Mole concept first derived from carbon atoms.