Question

What mass of a 0.583 molar solution of iron(III) nitrate is needed to obtain a) 0.0200 moles of iron(III) nitrate, b) 0.0500 moles of Fe^3+ ions, c) 0.00300 moles of nitrate ions? Please show work.

A) Mass = 0.583 g
B) Mass = 1.457 g
C) Mass = 0.0875 g
D) Mass = 0.872 g

Answer 1

The masses of a 0.583 molar solution of iron(III) nitrate needed to obtain specific moles of substance are 34.3 g for 0.0200 moles of iron(III) nitrate, 85.8 g for 0.0500 moles of Fe³⁺ ions and 1.716 g for 0.00300 moles of nitrate ions.

Step-by-step explanation:

To determine the mass of a 0.583 molar solution of iron(III) nitrate needed to obtain various amounts of iron(III) nitrate and its ions, we'll use molarity and stoichiometry.

a) To obtain 0.0200 moles of iron(III) nitrate:
Molarity (M) = moles of solute / volume of solution (L)
0.583 M = 0.0200 moles / V
V = 0.0200 moles / 0.583 M
V = 0.0343 L

Now, to find the mass, we need the density of the solution, but since it's not provided, we'll assume the density is similar to water (1 g/mL) for the purpose of this demonstration:

Mass = Volume × Density = 0.0343 L × 1000 mL/L × 1 g/mL = 34.3 g

b) The mole ratio of iron(III) to iron(III) nitrate is 1:1, therefore 0.0500 moles of Fe³⁺ ions is also 0.0500 moles of iron(III) nitrate:

0.583 M = 0.0500 moles / V
V = 0.0500 moles / 0.583 M
V = 0.0858 L
Mass = 0.0858 L × 1000 mL/L × 1 g/mL = 85.8 g

c) Each mole of iron(III) nitrate gives 3 moles of nitrate ions, so 0.00300 moles of nitrate ions equals 0.00100 moles of iron(III) nitrate:

0.583 M = 0.00100 moles / V
V = 0.00100 moles / 0.583 M
V = 0.001716 L
Mass = 0.001716 L × 1000 mL/L × 1 g/mL = 1.716 g