Calculate the weight of \[2.5\;\]moles of ethane (${C_2}{H_6}$) . The molecular weight of $C = 12{\text{g/mole}}$, $H = 1{\text{g/mole}}$
$75g$
$25g$
$30g$
$60g$

Hint: In this question, we shall calculate the molar mass of ${C_2}{H_6}$ . Then, simply use the unitary method to find the answer to this question. \[6.022 \times {10^{23}}\] is Avogadro's number and represents the number of atoms/molecules present in one mole of the substance. We shall use the mole concept to find the mass of the given moles of ${C_2}{H_6}$
 Formula used: ${\text{moles = }}\dfrac{{{\text{mass}}}}{{{\text{molar mass}}}}$

Complete step by step answer:
Ethane is a colourless and odourless liquid. It is easily inflammable and its vapours can cause asphyxiation. Frostbite can be caused by direct contact. It is used in the manufacturing of various chemicals.
 In the question, we are given the total number of moles. To find the total mass present we just need to multiply this value with a molar mass of ethane. Ethane contains two carbon atoms each weighing $12{\text{g/mole}}$ and six hydrogen atoms each weighing ${\text{1g/mole}}$
The molar mass of ethane = Mass of 1 mole of ethane = $2 \times 12 + 6 \times 1 = 24 + 6 = 30{\text{g}}$.
Mass of \[2.5\;\]moles of ethane= $2.5 \times 30 = 75{\text{g}}$.

Hence, the correct option is option A.

Note:
Avogadro constant can be defined as the number of molecules, atoms, or ions in one mole of a substance: $6.022 \times {10^{23}}$ per mol. It is derived from the number of atoms of the pure isotope $^{{\text{12}}}{\text{C}}$in 12 grams of that substance and is the reciprocal of atomic mass in grams. The formulae for the mole concept can be summarized as:
${\text{No}}{\text{. of moles = }}\dfrac{{{\text{Mass of the Substance in grams}}}}{{{\text{Molar mass of a Substance}}}} = \dfrac{{{\text{Number of Atoms or Molecules}}}}{{6.022 \times {{10}^{23}}}}$.