Question

How much mass of aluminium can be obtained from 0.1mol of $A{l}_2{(S{O}_4)}_3$ ? [Atomic mass of Al$=27u$ ]
(A) 2.7g
(B) 5.4g
(C) 3.2g
(D) 7.8g

Answer 1

Hint: This question can be solved using the formula -
$$\text{number of moles = }{\displaystyle \frac{\text{given mass}}{\text{molar mass}}}$$
Number of moles of Al in $A{l}_2{(S{O}_4)}_3$ can be calculated by multiplying number of atoms of Al with number of moles of $A{l}_2{(S{O}_4)}_3$ compound.

Complete answer:
We already know that a mole is defined as $$6.02214076\times {10}^{23}$$ (Avogadro’s number) of some chemical unit, be it atoms, molecules, ions, compounds or others.
Number of moles of $A{l}_2{(S{O}_4)}_3$ given as 0.1 mole.
Number of moles of Al in $A{l}_2{(S{O}_4)}_3$ can be calculated by multiplying number of atoms of Al with number of moles of $A{l}_2{(S{O}_4)}_3$ compound.
As 1 mole of $A{l}_2{(S{O}_4)}_3$ contains 2 moles of Al and 3 moles of $S{O}_4$ .
Similarly, 0.1 mole of $A{l}_2{(S{O}_4)}_3$ contains $(2\times 0.1=0.2)$0.2moles of Al and $(3\times 0.1=0.3)$ 0.3 mole of $S{O}_4$ .
We know that,
     $$\begin{array}{rl}& \text{number of moles = }{\displaystyle \frac{\text{given mass}}{\text{molar mass}}}\\ & \text{given mass}=\text{number of moles}\times \text{molar mass}\end{array}$$
Molar mass of aluminium is 27g/mol. And the number of moles of Aluminum is 0.2 (as calculated above). So putting these values we can find the mass.
     $$\begin{array}{rl}& \text{mass of Al = number of mole of Al }\times \text{ molar mass of Al}\\ & \text{mass of Al = }0.2\times 27\\ & \text{mass of Al =}5.4g\end{array}$$
Therefore, 5.4g of aluminium can be obtained from 0.1mol of $A{l}_2{(S{O}_4)}_3$

Hence the correct option is (B) 5.4g.

Note:
Molar mass is defined as the mass of one mole of chemical species. Whereas Atomic mass is the mass of one individual unit of that chemical species. The atomic mass in amu (atomic mass unit) of a substance is numerically equivalent to the mass in grams of one mole of that substance.