What mass of NaOH is required to precipitate all the Fe2+ ions from 50.0 mL of 0.200 M Fe(NO3)2 solution?

Fe+2 + 2NaOH ==> Fe(OH)2 + 2Na^+

How many moles Fe^+2 do you have? M x L = moles. So look at the equation and determine the moles of NaOH required to react with this many moles iron(II). Then grams NaOH = moles x molar mass.

To determine the mass of NaOH required for the precipitation of Fe2+ ions, we first need to understand the reaction between Fe2+ ions and NaOH.

The balanced chemical equation for the reaction is:

Fe(NO3)2 + 2NaOH → Fe(OH)2 + 2NaNO3

From the equation, we can see that one mole of Fe(NO3)2 reacts with two moles of NaOH to form one mole of Fe(OH)2.

Given:
- Volume of Fe(NO3)2 solution = 50.0 mL
- Concentration of Fe(NO3)2 solution = 0.200 M

Step 1: Convert volume to liters
The volume of the Fe(NO3)2 solution is given in milliliters (mL). We need to convert it to liters (L) for ease of calculation.
1 L = 1000 mL

50.0 mL ÷ 1000 = 0.050 L

Step 2: Calculate the number of moles of Fe(NO3)2
To calculate the number of moles, we use the formula:

moles = concentration (M) × volume (L)

moles of Fe(NO3)2 = 0.200 M × 0.050 L
moles of Fe(NO3)2 = 0.010 mol

Step 3: Calculate the number of moles of NaOH required
From the balanced chemical equation, it is clear that the molar ratio of Fe(NO3)2 to NaOH is 1:2. This means that for the complete reaction, 1 mole of Fe(NO3)2 reacts with 2 moles of NaOH.

Therefore, the number of moles of NaOH required is twice the number of moles of Fe(NO3)2.

moles of NaOH = 2 × moles of Fe(NO3)2
moles of NaOH = 2 × 0.010 mol
moles of NaOH = 0.020 mol

Step 4: Calculate the mass of NaOH required
To calculate the mass of NaOH required, we need to use the molar mass of NaOH.

The molar mass of NaOH is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol)
Molar mass of NaOH = 40.00 g/mol

mass of NaOH = moles of NaOH × molar mass of NaOH
mass of NaOH = 0.020 mol × 40.00 g/mol
mass of NaOH = 0.80 g

Therefore, 0.80 grams of NaOH is required to precipitate all the Fe2+ ions from the 50.0 mL of 0.200 M Fe(NO3)2 solution.

To find the mass of NaOH required to precipitate all the Fe2+ ions from the given Fe(NO3)2 solution, we need to determine the stoichiometry of the reaction between Fe(NO3)2 and NaOH.

The balanced equation for the reaction is:
Fe(NO3)2 + 2NaOH → Fe(OH)2 + 2NaNO3

From the balanced equation, we can see that 1 mole of Fe(NO3)2 reacts with 2 moles of NaOH to form 1 mole of Fe(OH)2.

First, we need to find the number of moles of Fe(NO3)2 in the 50.0 mL of 0.200 M solution:
Moles of Fe(NO3)2 = Volume (in L) × Concentration (in mol/L)
Moles of Fe(NO3)2 = 0.050 L × 0.200 mol/L
Moles of Fe(NO3)2 = 0.010 mol

Since the stoichiometry of the reaction is 1:2 (Fe(NO3)2:NaOH), we can determine the number of moles of NaOH required to completely react with the Fe(NO3)2:
Moles of NaOH = 2 × Moles of Fe(NO3)2
Moles of NaOH = 2 × 0.010 mol
Moles of NaOH = 0.020 mol

Now, we can calculate the mass of NaOH required using its molar mass. The molar mass of NaOH is:
Na: 22.99 g/mol
O: 16.00 g/mol
H: 1.01 g/mol

Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
Molar mass of NaOH = 39.99 g/mol

Mass of NaOH = Moles of NaOH × Molar mass of NaOH
Mass of NaOH = 0.020 mol × 39.99 g/mol
Mass of NaOH = 0.7998 g

Therefore, the mass of NaOH required to precipitate all the Fe2+ ions from the 50.0 mL of 0.200 M Fe(NO3)2 solution is approximately 0.7998 grams.