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Answer: The percentage of potassium iodide in the sample is 2.63 %.
Explanation:
The chemical equation for the reaction of iodide ions with bromine gas follows:
[tex]I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+[/tex] (i)
The chemical equation for the reaction of iodate ions with barium ions follows:
[tex]Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2[/tex] ......(ii)
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of barium iodate = 0.0596 g
Molar mass of barium iodate = 487.13 g/mol
Using equation 1:
[tex]\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles[/tex]
By Stoichiometry of the reaction (ii):
1 mole of barium iodate is produced by 2 moles of iodate ions
So, [tex]1.22\times 10^{-4} moles[/tex] of barium iodate will be produced by [tex]\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodate ions
By the stoichiometry of the reaction (i):
1 mole of iodate ions are produced by 1 moles of iodine ions
So, [tex]2.44\times 10^{-4} moles[/tex] of iodate ions will be produced by [tex]\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles[/tex] of iodine ions
Moles of potassium iodide = Moles of iodide ions = [tex]2.44\times 10^{-4}[/tex]
Since, the molar mass of potassium iodide = 166 g/mol
Using equation 1:
[tex]\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g[/tex]
To calculate the percentage by mass of a substance, the equation used is:
[tex]\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100[/tex]
Mass of a solution = 1.54 g
Mass of potassium iodide = 0.0405 g
Using above equation:
[tex]\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%[/tex]
Hence, the percentage of potassium iodide in the sample is 2.63 %.
Expressing the results of potassium iodide in percentage = 2.63%
The chemical reaction of iodine ions with Bromine gas can be expressed as :
I⁻ + 3Br₂ + 3H₂O -- > 6Br⁻ + IO₃ + 6H⁺ ----- ( 1 )
Chemical reaction between iodate ions with barium ions can be expressed as : Ba²⁺ + 2IO⁻₃ ------> Ba ( IO₃ )₂ --------- ( 2 )
Step 1 : Calculate the number of Barium iodate moles
mass of Barium iodate = 0.0596 g
molar mass of Barium iodate = 487.13 g/mol
from equation ( 1 )
moles of Barium iodate = ( 0.0596 ) / ( 487.13 ) = 1.22 * 10⁻⁴ moles
also from equation ( 1 ) the moles of potassium iodide = moles of iodide ions
= 2.44 * 10⁻⁴
molar mass of potassium iodide = 166 g/mol
Next step : Determine the mass of potassium iodide
moles of potassium * molar mass
= 2.44 * 10⁻⁴ * 166 g/mol = 0.0405 g
Final step : Determine the percentage of potassium iodide in the solution
Percentage = ( mass of potassium iodide / mass of solution ) * 100
= ( 0.0405 / 1.54 ) * 100
= 2.63%
Hence we can conclude that potassium iodide in percentage = 2.63%
Learn more : https://brainly.com/question/3388761
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