Answer:
[tex]m_{NH_3}=0.0213gNH_3[/tex]
Explanation:
Hello there!
In this case, since it is known that the reaction between ammonia and acetic acid is:
[tex]CH_3COOH+NH_3\rightarrow CH_3COONH_4[/tex]
It is possible for us to realize that the mole ratio of acetic acid (vinegar) to ammonia is 1:1, that is why we can relate the concentrations as follows:
[tex]M_{acid}V_{acid}=M_{base}V_{base}\\\\[/tex]
In such a way, by knowing that the volume of these two are the same, we infer that their concentrations is also de same; and therefore, the mass of ammonia is calculated as:
[tex]m_{NH_3}=0.005L*0.25\frac{molNH_3}{L}*\frac{17.04gNH_3}{1molNH_3}\\\\m_{NH_3}=0.0213gNH_3[/tex]
Regards!
