Question

# 1. Using the mass percent of iron in iron (III) chloride, calculate the number of moles...

1. Using the mass percent of iron in iron (III) chloride, calculate the number of moles of iron in 139.87g of FeCl3. ( 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?

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

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

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