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 Fe³+ ions,
c) 0.00300 moles of nitrate ions?

a) 6.44 g
b) 9.78 g
c) 1.75 g
d) 2.91 g

Answer 1

To obtain 0.0200 moles of iron(III) nitrate, you would need 9.78 g of a 0.583 molar solution of iron(III) nitrate. To obtain 0.0500 moles of Fe³+ ions, you would need 7.03 g of the solution. To obtain 0.00300 moles of nitrate ions, you would need 0.94 g of the solution.

Step-by-step explanation:

To calculate the mass of a 0.583 molar solution of iron(III) nitrate needed to obtain:

a) 0.0200 moles of iron(III) nitrate:

We can use the formula: Molarity = Moles of solute / Volume of solution in liters

0.583 M = 0.0200 mol / Volume

Volume = 0.0200 mol / 0.583 M = 0.0343 L

The mass of the solution can be calculated using the formula: Mass = Molarity * Volume * Molar mass

Mass = 0.583 M * 0.0343 L * 241.9 g/mol

Mass = 9.78 g

b) 0.0500 moles of Fe³+ ions:

Fe³+ ions are 1:1 with iron(III) nitrate, so we need 0.0500 moles of iron(III) nitrate.

Mass = Molarity * Volume * Molar mass = (0.583 M * 0.0500 mol) * 241.9 g/mol = 7.03 g

c) 0.00300 moles of nitrate ions:

Nitrate ions are 1:3 with iron(III) nitrate, so we need 0.00900 moles of iron(III) nitrate.

Mass = Molarity * Volume * Molar mass = (0.583 M * 0.00900 mol) * 241.9 g/mol = 0.94 g