Question

II. 50mL of FeCl2 solution 0.02mol.L-1 are added to 100mL of NaOH solution 0.03mol.L-1.

1. Write the chemical equation and then the net ionic equation.
2. Calculate the mass of the precipitate formed (Fe(OH)2). (Fe=56
H=1 O=16)
please help ​

Answer 1

(a)
`Fe^(2+)+2OH^-` →
`Fe(OH)_2`

(b) 0.09 g

Step-by-step explanation:

(a) Let's first write the original chemical equation:

`FeCl_2+2NaOH` →
`Fe(OH)_2+2NaCl`

The net ionic equation is when we break up all the substances that can be broken up into ions. From part (b), we know that the solid is
`Fe(OH)_2`, so this cannot be broken up into ions. Everything else can so:

`Fe^(2+)+2Cl^-+2Na^++2OH^-` →
`Fe(OH)_2+2Na^++2Cl^-`

Cross off the spectator ions, which are the ions that appear on both sides, and we're left with:

`Fe^(2+)+2OH^-` →
`Fe(OH)_2`

(b) We need to figure out the limiting reactant: is it iron or is it hydroxide?

The number of moles of Fe 2+ is:

`50mL*(0.02mol)/(1000mL) =0.001molFe^(2+)`

The number of moles of OH- is:

`100mL*(0.03mol)/(1000mL) =0.003molOH^-`

0.001 moles of iron will make 0.001 moles of Fe(OH)2 because they are in a 1:1 ratio in the chemical equation.

0.003 moles of hydroxide will make 0.0015 moles of Fe(OH)2 because the ratio of Fe(OH)2 to OH- is 1:2.

We now know that the iron is the limiting reactant and that the number of moles of precipitate formed is 0.001 moles.

Convert moles to grams using the molar mass of Fe(OH)2, which is 56 + 2 * 16 + 2 * 1 = 90 g/mol:

`0.001molFe(OH)_2*(90g)/(1mol) =0.09g`

So there are 0.09 grams of precipitate.

Answer 2

1.

FeCl₂ (aq) + 2NaOH (aq) -> Fe(OH)₂ (s) + 2NaCl (aq)

Fe²⁺ (aq) + 2OH⁻ (aq) = Fe(OH)₂

2. 0.09 g FeOH₂

Step-by-step explanation:

1.

Chemical equation:

FeCl₂ (aq) + NaOH (aq) -> Fe(OH)₂ (s) + NaCl (aq)

Don't forget to balance.

FeCl₂ (aq) + 2NaOH (aq) -> Fe(OH)₂ (s) + 2NaCl (aq)

Net ionic equation:

Remember, this means we take out all the spectator ions that aren't directly involved in the reaction. Only include those ions that formed a precipitate, (which is the solid), and include their states of matter as well.

Fe²⁺ (aq) + 2OH⁻ (aq) = Fe(OH)₂

I know that Fe has a +2 charge, not a +3 charge, because it's paired with Cl2, which is a -2 charge, and the charges must balance to equal 0.

Always check again to make sure it's balanced even in net ionic form!

2. Now we have to get the mass of Fe(OH)₂, iron (II) hydroxide. First, add together the molar masses and multiply by how many there are to get the molar mass:

Fe: 56 g/mol * 1 = 56

O: 16 g/mol * 2 = 32

H: 1 g/mol * 2 = 2

56 + 32 + 2 = 90 g/mol

Now we have to figure out the actual mass, using the molar mass and the moles to get there. First solve for the moles, because we were given the concentration and the volume.

50 mL FeCl₂ = .050 L FeCl₂

.050 L FeCl₂ x .02 mol/L FeCl₂ = .001 mol FeCl₂

100 mL NaOH = .1 L NaOH

.1 L NaOH x .03 mol/L = .003 mol NaOH

To find out which is the limiting reactant, refer back to our chemical equation.

FeCl₂ (aq) + 2NaOH (aq) -> Fe(OH)₂ (s) + 2NaCl (aq)

For every 1 mol of FeCl₂, we need 2 mol NaOH. So let's start with FeCl₂ being the limiting reactant:

If we have .001 mol FeCl₂ and we use all of it, we would need .002 mol NaOH. We have that. So FeCl₂ must be the limiting reactant.

If we have .001 mol FeCl₂, we would produce .001 mol FeOH₂, according to the equation. Now let's multiply that by the molar mass to find the mass of the precipitate in grams.

.001 mol FeOH₂ x 90 g/mol = .09 g FeOH₂