Question

1. Using the mass percent of iron in iron (III) chloride,
calculate the number of moles of iron in 139.87g of
FeCl_{3}. ( The Molar mass of iron is 55.85 grams.)

2. If 2.317 g of Iron (III) bromide is transferred to a volumetric flask, what is the molarity (M), if the volume of solution is 250 ml?

If 2.317 g of Iron (III) bromide is transferred to a volumetric flask, what is the molarity (M), if the volume of solution is 250 ml?

Answer #1

1)

Molar mass of FeCl3,

MM = 1*MM(Fe) + 3*MM(Cl)

= 1*55.85 + 3*35.45

= 162.2 g/mol

MM of FeCl3 = 162.2

Lets calculate the mass of element

mass of Fe = 1*MM(Fe)

= 55.85 g

Mass % of Fe = (55.85*100)/162.2

= 34.43 %

SO,

mass of Fe = 34.43 % of 139.87

= 34.43 * 139.87 / 100

= 48.16 g

Molar mass of Fe = 55.85 g/mol

mass(Fe)= 48.16 g

use:

number of mol of Fe,

n = mass of Fe/molar mass of Fe

=(48.16 g)/(55.85 g/mol)

= 0.8623 mol

Answer: 0.8623 mol

2)

Molar mass of FeBr3,

MM = 1*MM(Fe) + 3*MM(Br)

= 1*55.85 + 3*79.9

= 295.55 g/mol

mass(FeBr3)= 2.317 g

use:

number of mol of FeBr3,

n = mass of FeBr3/molar mass of FeBr3

=(2.317 g)/(295.55 g/mol)

= 7.84*10^-3 mol

volume , V = 250 mL

= 0.25 L

use:

Molarity,

M = number of mol / volume in L

= 7.84*10^-3/0.25

= 3.136*10^-2 M

Answer: 3.136*10^-2 M

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