Complete Question
The complete question is shown on the first and second uploaded image
Answer:
A
The percentage of water of hydration is [tex]P_h= 11.01[/tex]%
Mass of [tex]Fe^{3+}[/tex] in 100mg is 10.60mg
Moles of [tex]Fe^{3+}[/tex] in 100mg is [tex]n_i= 0.19[/tex]
mol / mol Fe (3 sig figs) is [tex]= 1.00[/tex]
mol / mol Fe (whole number) is = 1
B
Mass of [tex]K^{+}[/tex] in 100mg is 27.70mg
Moles of [tex]K^{+}[/tex] in 100mg is [tex]n_i= 0.581 moles[/tex]
mol of K / mol of Fe (3 sig figs) is [tex]= 3.05[/tex]
mol of K / mol of Fe (whole number) is [tex]=3[/tex]
C
Mass of [tex]C_2O_4^{-2}[/tex] in 100mg is 55.69 mg
Moles of [tex]C_2O_4^{-2}[/tex] in 100mg is [tex]n_i= 0.633 moles[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= 3.33[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (whole number) is [tex]=3[/tex]
D
Mass of water in 100mg is 11.01 mg
Moles of water in 100mg is [tex]n_i= 0.611 moles[/tex]
mol of water / mol of Fe (3 sig figs) is [tex]= 3.21[/tex]
mol of water / mol of Fe (whole number) is [tex]=3[/tex]
Explanation:
The percentage of water of hydration is mathematically represented as
[tex]P_h = 100 - (Pi + P_p + P_o)[/tex]
Now substituting 10.60% for [tex]P_i[/tex] (percentage of iron ) , 22.70% for [tex]P_p[/tex](Percentage of potassium) , 55.69% for [tex]P_o[/tex] (percentage of Oxlate)
[tex]P_h =100 - (10.60 + 22.70+55.69)[/tex]
[tex]P_h= 11.01[/tex]%
For IRON
Since the percentage of [tex]Fe^{3+}[/tex] is 10.60% then in a 100 mg of the sample the amount of [tex]Fe^{3+}[/tex] would be 10.60 mg
Now the no of moles is mathematically denoted as
[tex]n = \frac{mass}{molar \ mass }[/tex]
The molar mass of [tex]Fe[/tex] is 55.485 g/mol
So the number of moles of [tex]Fe^{3+}[/tex] in 100mg of he sample is
[tex]n_i = \frac{10.60}{55.485}[/tex]
[tex]n_i= 0.19[/tex]
mol / mol Fe (3 sig figs) is [tex]= \frac{0.19}{0.19} = 1.00[/tex]
FOR POTASSIUM
Since the percentage of [tex]K^{+}[/tex] is 22.70% then in a 100mg of the sample the amount of [tex]K^{+}[/tex] would be 22.70mg
The molar mass of [tex]K[/tex] is 39.1 g/mol
So the number of moles of [tex]K^{+}[/tex] in 100mg of he sample is
[tex]n_i = \frac{22.70}{39.1}[/tex]
[tex]=0.581 moles[/tex]
mol of K / mol of Fe (3 sig figs) is [tex]= \frac{0.581}{0.19} = 3.05[/tex]
FOR OXILATE [tex]C_2O_4^{-2}[/tex]
Since the percentage of [tex]C_2O_4^{-2}[/tex] is 55.69% then in a 100mg of the sample the amount of [tex]C_2O_4^{-2}[/tex] would be 55.69 mg
The molar mass of [tex]C_2O_4^{-2}[/tex] is 88.02 g/mol
So the number of moles of [tex]C_2O_4^{-2}[/tex] in 100mg of he sample is
[tex]n_i = \frac{55.69}{88.02}[/tex]
[tex]=0.633 moles[/tex]
mol of [tex]C_2O_4^{-2}[/tex] / mol of Fe (3 sig figs) is [tex]= \frac{0.633}{0.19} = 3.33[/tex]
FOR WATER OF HYDRATION
Since the percentage of water is 11.01% then in a 100mg of the sample the amount of water would be 11.0 mg
The molar mass of water is 18.0 g/mol
So the number of moles of water in 100mg of he sample is
[tex]n_i = \frac{11.01}{18.0}[/tex]
[tex]=0.611 moles[/tex]
mol of water / mol of Fe (3 sig figs) is [tex]= \frac{0.611}{0.19} = 3.21[/tex]
