A student determines the iron (II) content of a solution by first precipitating it as iron (II) hydroxide, and then decomposing the hydroxide to iron (II) oxide by heating. How many grams of iron (II) oxide should the student obtain if her solution contains 46.0 mL of 0.426 M iron(II) nitrate?

Fe(NO3)2 + 2OH^- ==> Fe(OH)2 + 2NO3^-

Then Fe(OH)2 + heat ==> FeO + H2O
mols Fe(NO3)2 = mols x L = ?
mols FeO = mols Fe(NO3)2 (look at the equation; 1 mol Fe(NO3)2 = 1 mol FeO)
Then g Fe) = mols FeO x molar mass FeO.

To determine the grams of iron (II) oxide, we need to set up a series of steps. Let's break it down:

1. Determine the moles of iron (II) nitrate in the solution:
Moles = concentration (M) x volume (L)
Moles = 0.426 M x 0.0460 L
Moles = 0.019596 mol

2. Determine the moles of iron (II) hydroxide formed:
From the balanced chemical equation for the precipitation of iron (II) hydroxide:
Fe(NO3)2 + 2NaOH -> Fe(OH)2 + 2NaNO3
1 mole of Fe(NO3)2 produces 1 mole of Fe(OH)2
Moles of Fe(OH)2 = 0.019596 mol

3. Determine the moles of iron (II) oxide formed:
From the balanced chemical equation for the decomposition of iron (II) hydroxide to iron (II) oxide:
2Fe(OH)2 -> 2FeO + 2H2O
1 mole of Fe(OH)2 produces 1 mole of FeO
Moles of FeO = 0.019596 mol

4. Determine the molar mass of iron (II) oxide (FeO):
The molar mass of iron (II) oxide is the sum of the atomic masses of iron (Fe) and oxygen (O):
Molar mass of Fe = 55.845 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of FeO = 55.845 g/mol + 16.00 g/mol
Molar mass of FeO = 71.845 g/mol

5. Determine the grams of iron (II) oxide obtained:
Grams = moles x molar mass
Grams = 0.019596 mol x 71.845 g/mol
Grams = 1.4097 g

Therefore, the student should obtain approximately 1.41 grams of iron (II) oxide.

To determine the grams of iron (II) oxide obtained, you need to follow a series of steps:

Step 1: Calculate the number of moles of iron (II) nitrate.
Given that the volume of the solution is 46.0 mL and its concentration is 0.426 M (moles per liter), we can calculate the number of moles using the formula:

Moles = Volume (in liters) × Concentration

Converting the volume from milliliters to liters:
46.0 mL ÷ 1000 mL/L = 0.046 L

Calculating the number of moles:
Moles = 0.046 L × 0.426 moles/L = 0.019596 moles

Step 2: Determine the stoichiometry between iron (II) nitrate and iron (II) oxide.
From the balanced chemical equation, we can see that for every 1 mole of iron (II) nitrate, we obtain 1 mole of iron (II) oxide.

Step 3: Convert moles of iron (II) nitrate to moles of iron (II) oxide.
Since the stoichiometry ratio between the two compounds is 1:1, the number of moles of iron (II) oxide will be the same as the number of moles of iron (II) nitrate, which is 0.019596 moles.

Step 4: Calculate the molar mass of iron (II) oxide.
The molar mass of any compound is the sum of the atomic masses of all the elements present. In this case, iron (II) oxide (FeO) contains iron (Fe) and oxygen (O).

Molar mass(Fe) = 55.85 g/mol
Molar mass(O) = 16.00 g/mol

Molar mass(FeO) = Molar mass(Fe) + Molar mass(O)
= 55.85 g/mol + 16.00 g/mol
= 71.85 g/mol

Step 5: Calculate the mass of iron (II) oxide.
Mass = Moles × Molar mass

Mass = 0.019596 moles × 71.85 g/mol
≈ 1.41 grams

Therefore, the student should obtain approximately 1.41 grams of iron (II) oxide from the solution.