### Theory:

 1 Calculate the volume occupied by $$2.5$$ $$moles$$ of $$CO_2$$ at STP.

Data:

Given number of moles of $$CO_2=2.5$$

Molar volume of $$CO_2$$ at STP $$=22.4$$ $$L$$

Solution:

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}& =& \frac{\mathit{Volume}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}\phantom{\rule{0.147em}{0ex}}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}}{\mathit{Molar}\phantom{\rule{0.147em}{0ex}}\mathit{volume}\phantom{\rule{0.147em}{0ex}}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}}\\ 2.5& =& \frac{\mathit{Volume}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}\phantom{\rule{0.147em}{0ex}}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}}{22.4}\\ \mathit{Volume}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}\phantom{\rule{0.147em}{0ex}}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}& =& 22.4×2.5\\ & =& 56\phantom{\rule{0.147em}{0ex}}L\end{array}$

Therefore, the volume occupied by $$2.5$$ $$mole$$ of $$CO_2$$ at STP $$=56$$ $$L$$

 2 Calculate the volume occupied by $$12.046\times10^{23}$$ of ammonia gas molecules.

Data:

Given the number of ammonia gas molecules $$=12.046\times10^{23}$$

Avogadro’s number $$=6.023\times10^{23}$$

Solution:

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}& =& \frac{\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{molecules}}{\mathit{Avogadro}"s\phantom{\rule{0.147em}{0ex}}\mathit{number}}\\ & =& \frac{12.046×{10}^{23}}{6.023×{10}^{23}}\\ & =& 2\phantom{\rule{0.147em}{0ex}}\mathit{moles}\end{array}$

$\begin{array}{lll}\mathit{Volume}\phantom{\rule{0.147em}{0ex}}\mathit{occupied}\phantom{\rule{0.147em}{0ex}}\mathit{by}\phantom{\rule{0.147em}{0ex}}{\mathit{NH}}_{3}& =& \mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}×\mathit{Molar}\phantom{\rule{0.147em}{0ex}}\mathit{volume}\\ & =& 2×22.4\\ & =& 44.8\phantom{\rule{0.147em}{0ex}}L\end{array}$
Therefore, the volume occupied by $$12.046\times10^{23}$$ of ammonia gas molecules $$=44.8$$ $$L$$

 3 Calculate the volume occupied by $$14$$ $$g$$ of nitrogen gas.

Data:

Given mass of nitrogen $$=14$$ $$g$$

Atomic mass of nitrogen $$=28$$

Solution:

$\begin{array}{lll}\mathit{Number}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}& =& \frac{\mathit{Given}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{element}}{\mathit{Atomic}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{the}\phantom{\rule{0.147em}{0ex}}\mathit{element}}\\ & =& \frac{14}{28}\\ & =& 0.5\phantom{\rule{0.147em}{0ex}}\mathit{mole}\end{array}$

$\begin{array}{lll}\mathit{Volume}\phantom{\rule{0.147em}{0ex}}\mathit{occupied}\phantom{\rule{0.147em}{0ex}}\mathit{by}\phantom{\rule{0.147em}{0ex}}{N}_{2}\mathit{at}\phantom{\rule{0.147em}{0ex}}\mathit{STP}& =& \mathit{No}.\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}\mathit{moles}×\mathit{Molar}\phantom{\rule{0.147em}{0ex}}\mathit{volume}\\ \mathit{Mass}& =& 0.5×22.4\\ & =& 11.2\phantom{\rule{0.147em}{0ex}}L\end{array}$
Therefore, the volume occupied by $$14$$ $$g$$ of nitrogen gas $$=11.2$$ $$L$$