Theory:

 1 Calculate the mass of $$0.3$$ $$mole$$ of aluminium (atomic mass of $$Al=27$$).

Data:

Given moles of aluminium $$= 0.3$$ $$mole$$

Atomic mass of aluminium $$= 27$$

Formula:

$\begin{array}{l}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}=\frac{\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Al}}{\mathit{Atomic}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Al}}\\ \mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Al}=\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}×\mathit{Atomic}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Al}\end{array}$

Solution:

By substituting the values to the above formula, we get,

$\begin{array}{lll}\mathit{Mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{Al}& =& 0.3×27\\ & =& 8.1\phantom{\rule{0.147em}{0ex}}g\end{array}$

Hence, the mass of $$0.3$$ $$mole$$ of aluminium $$=8.1$$ $$g$$

 2 Calculate the mass of $$2.24$$ $$litre$$ of $$SO_2$$ gas at STP.

Data:

Given volume of $$SO_2=2.24$$ $$L$$

Atomic mass of $$S=32$$

Atomic mass of $$O=16$$

Solution:

$\begin{array}{lll}\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{SO}}_{2}& =& 32+\left(16×2\right)\\ & =& 32+32\\ & =& 64\end{array}$

$\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{SO}}_{2}=\frac{\mathit{Given}\phantom{\rule{0.147em}{0ex}}\mathit{volume}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{SO}}_{2}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}\phantom{\rule{0.147em}{0ex}}}{\mathit{Molar}\phantom{\rule{0.147em}{0ex}}\mathit{volume}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{SO}}_{2}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}\phantom{\rule{0.147em}{0ex}}}$

By substituting the values in the above formula, we get,

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{SO}}_{2}& =& \frac{2.24}{22.4}\\ & =& 0.1\phantom{\rule{0.147em}{0ex}}\mathit{mole}\end{array}$

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}& =& \frac{\mathit{Mass}}{\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}}\\ \mathit{Mass}& =& \mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}×\mathit{Molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\\ & =& 0.1×64\\ & =& 6.4\phantom{\rule{0.147em}{0ex}}g\end{array}$

Hence, the mass of $$2.24$$ $$litre$$ of $$SO_2=6.4$$ $$g$$