Answer:
1.
FeCl₂ (aq) + 2NaOH (aq) -> Fe(OH)₂ (s) + 2NaCl (aq)
Fe²⁺ (aq) + 2OH⁻ (aq) = Fe(OH)₂
2. 0.09 g FeOH₂
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
