What mass of NaOH is required to precipitate all the Fe2+ ions from 52.0 mL of 0.200 M Fe(NO3)2 solution?
Write the equation.
Fe^+2 + 2OH^- ==> Fe(OH)2
moles Fe^+2 = M x L
moles NaOH must be 2 x moles Fe^+2.
moles OH is same as moles NaOH.
g NaOH = moles NaOH x molar mass NaOH.
Ah, chemistry, the art of turning equations into a comedic tragedy! Now, let's see what we have here. To calculate the mass of NaOH needed, we need to determine the moles of Fe2+ in the solution first.
So, we have 52.0 mL of 0.200 M Fe(NO3)2 solution. Let's convert this into moles of Fe2+.
52.0 mL is like saying "I'm 52.0 milliliters cool!" To convert this into liters, we just divide by 1000 and get 0.052 L.
Now, we can calculate the moles of Fe2+ with the equation:
moles = volume (in liters) x concentration
moles of Fe2+ = 0.052 L x 0.200 moles/L
Wow, this is exciting! Calculations make me laugh so hard! So, let's do the math:
moles of Fe2+ = 0.0104 moles
Now comes the serious part. We need to ensure that all the Fe2+ ions precipitate by reacting with NaOH. The balanced equation for this reaction is:
2 NaOH + Fe(NO3)2 → Fe(OH)2 + 2 NaNO3
From the equation, we see that 1 mole of Fe(NO3)2 reacts with 2 moles of NaOH. Since the stoichiometry demands it, we need twice as many moles of NaOH as we have of Fe2+.
So, we multiply the moles of Fe2+ by 2 to find the moles of NaOH needed:
moles of NaOH = (0.0104 moles of Fe2+) x 2 = 0.0208 moles
Now, to make things simpler, let's turn moles of NaOH into grams. But first, let me do a funny dance. *does a silly dance*
The molar mass of NaOH is approximately 40.00 g/mol. So, we can find the mass of NaOH needed by multiplying the moles by the molar mass:
mass of NaOH = 0.0208 moles x 40.00 g/mol
And when we do the calculation, we get:
mass of NaOH = 0.832 g
So, 0.832 grams of NaOH should be enough to precipitate all the Fe2+ ions in your 52.0 mL of 0.200 M Fe(NO3)2 solution. Hope this answer brings a smile to your face!
To find the mass of NaOH required to precipitate all the Fe2+ ions, we need to determine the stoichiometry of the reaction between Fe2+ ions and NaOH.
The balanced chemical equation for the reaction between Fe2+ ions and NaOH is:
2 Fe2+ (aq) + 2 OH– (aq) → Fe(OH)2 (s)
From the balanced equation, we can see that it requires 2 moles of Fe2+ ions to react with 2 moles of OH– ions to form 1 mole of Fe(OH)2 precipitate.
Step 1: Calculate the number of moles of Fe2+ ions in the solution.
Given:
Volume of Fe(NO3)2 solution = 52.0 mL = 0.0520 L
Concentration of Fe(NO3)2 solution = 0.200 M
Using the equation: moles = concentration × volume, we can calculate the number of moles of Fe2+ ions in the solution:
moles of Fe2+ ions = 0.200 M × 0.0520 L
moles of Fe2+ ions = 0.0104 moles
Step 2: Calculate the number of moles of NaOH required to react with the Fe2+ ions.
From the balanced equation, we know that it takes 2 moles of Fe2+ ions to react with 2 moles of OH– ions. Therefore, the number of moles of NaOH required is the same as the number of moles of Fe2+ ions.
moles of NaOH = 0.0104 moles
Step 3: Convert moles of NaOH to mass of NaOH.
To convert moles of NaOH to mass, we need to know the molar mass of NaOH.
The molar mass of NaOH = (sodium atomic mass) + (oxygen atomic mass) + (hydrogen atomic mass)
The molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol)
The molar mass of NaOH = 40.00 g/mol
Mass of NaOH = moles of NaOH × molar mass of NaOH
Mass of NaOH = 0.0104 moles × 40.00 g/mol
Mass of NaOH = 0.416 g
Therefore, 0.416 grams of NaOH is required to precipitate all the Fe2+ ions from the 52.0 mL of 0.200 M Fe(NO3)2 solution.
To determine the mass of NaOH required to precipitate all Fe2+ ions from the Fe(NO3)2 solution, we need to calculate the number of moles of Fe2+ ions present in the solution and use the stoichiometry of the balanced chemical equation to find the corresponding moles of NaOH.
1. First, let's calculate the number of moles of Fe2+ ions in the 0.200 M Fe(NO3)2 solution. We can use the formula:
Moles = Molarity * Volume (in liters)
Since the volume is given in milliliters, we need to convert it to liters by dividing by 1000:
Volume (in liters) = 52.0 mL / 1000 = 0.0520 L
Moles of Fe2+ ions = 0.200 M * 0.0520 L = 0.0104 moles
2. Next, we need to determine the stoichiometry between Fe2+ ions and NaOH from the balanced chemical equation. The balanced equation for the reaction of Fe2+ ions with NaOH 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 produce 1 mole of Fe(OH)2.
3. Using the stoichiometry, we can calculate the moles of NaOH required to react with 0.0104 moles of Fe2+ ions:
Moles of NaOH = 2 * Moles of Fe2+ ions
Moles of NaOH = 2 * 0.0104 moles = 0.0208 moles
4. Finally, we need to find the mass of NaOH required. We can do this by multiplying the moles of NaOH by its molar mass. The molar mass of NaOH is:
Molar mass of NaOH = 22.99 g/mol (from the atomic masses of sodium (Na), oxygen (O), and hydrogen (H))
Mass of NaOH = Moles of NaOH * Molar mass of NaOH
Mass of NaOH = 0.0208 moles * 22.99 g/mol = 0.478 g (rounded to three decimal places)
Therefore, approximately 0.478 grams of NaOH is required to precipitate all the Fe2+ ions from the 52.0 mL of 0.200 M Fe(NO3)2 solution.