Respuesta :
Answer: The mass percentage of nitrogen in the sample is 6.04 %
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
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:
[tex]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[/tex]
The chemical equation for the reaction of HCl and ammonia follows:
[tex]HCl+NH_3\rightarrow NH_4Cl[/tex]
By Stoichiometry of the reaction:
1 mole of HCl reacts with 1 mole of ammonia
So, [tex]9.75\times 10^{-3}mol[/tex] of HCl will react with = [tex]\frac{1}{1}\times 9.75\times 10^{-3}mo=9.75\times 10^{-3}mol[/tex] of ammonia
1 mole of ammonia contains 1 mole of nitrogen and 3 moles of hydrogen element
Moles of nitrogen in ammonia = [tex]9.75\times 10^{-3}mol[/tex]
- To calculate the mass for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Moles of nitrogen = [tex]9.75\times 10^{-3}mol[/tex]
Molar mass of nitrogen = 14 g/mol
Putting values in above equation, we get:
[tex]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[/tex]
- To calculate the mass percentage of nitrogen in the sample, we use the equation:
[tex]\text{Mass percent of nitrogen}=\frac{\text{Mass of nitrogen}}{\text{Mass of sample}}\times 100[/tex]
Mass of sample= 2.25 g
Mass of nitrogen = 0.136 g
Putting values in above equation, we get:
[tex]\text{Mass percent of nitrogen}=\frac{0.136g}{2.25g}\times 100=6.04\%[/tex]
Hence, the mass percentage of nitrogen in the sample is 6.04 %
