<u>Answer:</u> The mass percentage of nitrogen in the sample is 6.04 %
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Molarity of HCl solution = 0.150 M
Volume of solution = 65.0 mL = 0.065 L (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
![0.150M=\frac{\text{Moles of HCl}}{0.065L}\\\\\text{Moles of HCl}=(0.150mol/L\times 0.065L)=9.75\times 10^{-3}mol](https://tex.z-dn.net/?f=0.150M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20HCl%7D%7D%7B0.065L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20HCl%7D%3D%280.150mol%2FL%5Ctimes%200.065L%29%3D9.75%5Ctimes%2010%5E%7B-3%7Dmol)
The chemical equation for the reaction of HCl and ammonia follows:
![HCl+NH_3\rightarrow NH_4Cl](https://tex.z-dn.net/?f=HCl%2BNH_3%5Crightarrow%20NH_4Cl)
By Stoichiometry of the reaction:
1 mole of HCl reacts with 1 mole of ammonia
So,
of HCl will react with =
of ammonia
1 mole of ammonia contains 1 mole of nitrogen and 3 moles of hydrogen element
Moles of nitrogen in ammonia = ![9.75\times 10^{-3}mol](https://tex.z-dn.net/?f=9.75%5Ctimes%2010%5E%7B-3%7Dmol)
- To calculate the mass for given number of moles, we use the equation:
Moles of nitrogen = ![9.75\times 10^{-3}mol](https://tex.z-dn.net/?f=9.75%5Ctimes%2010%5E%7B-3%7Dmol)
Molar mass of nitrogen = 14 g/mol
Putting values in above equation, we get:
![9.75\times 10^{-3}mol=\frac{\text{Mass of nitrogen}}{14g/mol}\\\\\text{Mass of nitrogen}=(9.75\times 10^{-3}mol\times 14g/mol)=0.136g](https://tex.z-dn.net/?f=9.75%5Ctimes%2010%5E%7B-3%7Dmol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%7D%7D%7B14g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20nitrogen%7D%3D%289.75%5Ctimes%2010%5E%7B-3%7Dmol%5Ctimes%2014g%2Fmol%29%3D0.136g)
- To calculate the mass percentage of nitrogen in the sample, we use the equation:
![\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of sample}}\times 100](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20nitrogen%7D%3D%5Cfrac%7B%5Ctext%7BMass%20of%20nitrogen%7D%7D%7B%5Ctext%7BMass%20of%20sample%7D%7D%5Ctimes%20100)
Mass of sample= 2.25 g
Mass of nitrogen = 0.136 g
Putting values in above equation, we get:
![\text{Mass percent of nitrogen}=\frac{0.136g}{2.25g}\times 100=6.04\%](https://tex.z-dn.net/?f=%5Ctext%7BMass%20percent%20of%20nitrogen%7D%3D%5Cfrac%7B0.136g%7D%7B2.25g%7D%5Ctimes%20100%3D6.04%5C%25)
Hence, the mass percentage of nitrogen in the sample is 6.04 %