Respuesta :
Answer:
The mass% of iodide ions in the sample is 20.13%.
Explanation:
Total Moles of silver nitrate added = n
Molarity of the silver nitrate solution = 0.05539 M
Volume of the silver nitrate = 50.00mL = 0.050 L
[tex]Moles=Molarity\times Volume (L)[/tex]
[tex]n= 0.05539 M\times 0.050 L=0.0027695 mol[/tex]
[tex]AgNO_3+KSCN\rightarrow AgSCN+KNO_3[/tex]..[1]
Moles of potassium thiocyanate = n'
Molarity of the potassium thiocyanate solution = 0.05233 M
Volume of the potassium thiocyanate = 33.12 mL= 0.03312 L
[tex]n'= 0.05539 M\times 0.03312 L=0.0017332 mol[/tex]
According to reaction-1, 1 mole of potassium thiocyanate reacts with 1 mol of silver nitrate. Then 0.0017332 mole of potassium thiocyanate reacts with 0.0017332 mol of silver nitrate.
Moles of silver nitrate which had reacted with iodide ions = N
n = n' + N
N = n - n' = 0.0027695 mol - 0.0017332 = 0.0010363 mol
[tex]I^-+AgNO_3\rightarrow AgI+NO_{3}^-[/tex]..[2]
According to reaction-2, 1 mole of iodide reacts with 1 mol of silver nitrate. Then 0.0010363 mole of silver nitrate reacts with :
[tex]\frac{1}{1}0.0010363 mol= 0.0010363 mol[/tex] of silver nitrate.
Mass of 0.0010363 mole of iodide ions =
0.0010363 mol × 126.9 g/mol = 0.1315 g
The mass% of iodide ions in the sample:
Mass of sample = 0.6532 g
[tex]=\frac{\text{Mass of iodide ions}}{\text{mass of sample}}\times 100[/tex]
[tex]=\frac{0.1315 g}{0.6532g}\times 100=20.13\%[/tex]
