Passage type questions:

            1 mole of any gas occupies 22.4 litre volume at STP.

            Mass of a gas can be related to volume according to the following ideal gas equation. 

            PV = nRT                       R = Universal gas constant

            $\text{Number of moles (n)}=\frac{\text{Weight}}{\text{Molecular Weight}}$

                   $\text{PV}=\frac{\text{w}}{\text{m}}\text{RT}$ 

Q.  The volume ratio of ${\text{H}}_{\text{2}}$ gas and ${\text{N}}_{\text{2}}$  gas which are formed by the decomposition of 65 g of  ${\text{NH}}_{\text{3}}$ at STP     

${\text{2NH}}_3\to {\text{3H}}_{\text{2}}+{\text{N}}_{\text{2}}$

$2{\text{\hspace{0.17em}moles of NH}}_3\to {\text{ 3 moles of H}}_{\text{2}}{\text{and\hspace{0.17em}One\hspace{0.17em}mole\hspace{0.17em}of\hspace{0.17em}N}}_{\text{2}}$

$2\times 17g{\text{\hspace{0.17em}of NH}}_3\to 3\times 22.4{\text{\hspace{0.17em}Lts\hspace{0.17em}Volume of H}}_{\text{2}}\text{\hspace{0.17em}and }22.4{\text{\hspace{0.17em}Lts\hspace{0.17em}Volume of\hspace{0.17em}N}}_{\text{2}}$                          

 $34g{\text{\hspace{0.17em}of NH}}_3\to {\text{67.2\hspace{0.17em}Lts\hspace{0.17em}Volume of H}}_{\text{2}}\text{\hspace{0.17em}and }22.4{\text{\hspace{0.17em}Lts\hspace{0.17em}Volume of\hspace{0.17em}N}}_{\text{2}}$                                      

$34g{\text{\hspace{0.17em}of NH}}_3\to 67.2{\text{Lts\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}$                    $34g{\text{\hspace{0.17em}of NH}}_3\to 22.4{\text{\hspace{0.17em}Lts\hspace{0.17em}of\hspace{0.17em}N}}_{\text{2}}$    

$65g{\text{\hspace{0.17em}of NH}}_3\to ?\text{X}$                                        $65g{\text{\hspace{0.17em}of NH}}_3\to ?\text{X}$       

$\text{X=}\frac{\text{65×67.2}}{34}$                                               $\text{X=}\frac{\text{65×22.4}}{34}$                                                                                                 

             X=128.47 Lts of                                 X=42.82 Lts of 

Ratio of  ${\text{Volumes of H}}_{\text{2}}{\text{and\hspace{0.17em}N}}_{\text{2}}$ formed is 128.47: 42.82= 3:1

On decomposition of Ammonia releases 3 moles of Hydrogen and 1 mole of Nitrogen.