Calculate the number of millilitres of NH3(aq) solution (d = 0.986 g/ mL) contain 2.5 % by weight NH3 which will be required to precipitate iron as Fe(OH)3 in a 0.8 g sample that contains 50 % Fe2O3.
wt Fe2O3 = 0.8 x 1/2 = 0.4gram.
Convert 0.4g Fe2O3 to g Fe. That's
0.4g x (2 atomic mass Fe/molar mass Fe2O3) = estimated 0.28g Fe = estd 0.005 mol Fe (that's mol = g/atomic mass).
So how much OH^- is needed?
Fe^3+ + 3OH^- ==> Fe(OH)3
0.005 mols Fe^3+ will require 3*0.005 mols OH^- = about 0.015
What's the concentration of the NH3? That's
1000 mL x 0.0986 g/mL x 0.025 x (1/molar mass NH3) = estd 0.145M
Then M NH3 = mols NH3/L NH3. You have M and you have mols NH3 needed, solve for L NH3 needed and convert to mL.
Well, you've got quite the chemical circus going on there! Let's tackle this step by step, while keeping it fun, of course.
First, we need to find out how much Fe(OH)3 is present in the 0.8 g sample that contains 50% Fe2O3. So, 50% of 0.8 g is 0.4 g of Fe2O3.
Now, time to perform a balancing act! The balanced equation for the reaction between NH3 and Fe(OH)3 is:
2NH3 + 3Fe(OH)3 → Fe2O3 + 6H2O
So, according to the equation, you need 3 moles of Fe(OH)3 to react with 2 moles of NH3. Since we know the molar mass of Fe(OH)3 is approximately 106 g/mol, we can find out how many moles we have in 0.4 g.
0.4 g / 106 g/mol = approximately 0.0038 moles of Fe(OH)3
Since the ratio is 3 moles of Fe(OH)3 to 2 moles of NH3, we can calculate how many moles of NH3 will be required:
(0.0038 moles Fe(OH)3 x 2 moles NH3) / 3 moles Fe(OH)3 = approximately 0.00253 moles NH3
Finally, let's convert this number of moles into milliliters of NH3(aq). We need the density of NH3(aq), which is given as 0.986 g/mL.
Since we know the percentage by weight of NH3 in the solution is 2.5%, we can calculate the mass of NH3 in the solution as follows:
0.00253 moles NH3 x 17 g/mol (molar mass of NH3) = approximately 0.043 g NH3
Now, let's find the volume using the density:
0.043 g NH3 / 0.986 g/mL ≈ 0.0437 mL NH3(aq)
So, approximately 0.0437 milliliters of NH3(aq) solution will be required to precipitate the iron as Fe(OH)3 in the 0.8 g sample. And there you have it, a chemical balancing act with a hint of comedy!
To calculate the number of milliliters of NH3(aq) solution required to precipitate iron as Fe(OH)3, we need to follow these steps:
Step 1: Calculate the amount of Fe2O3 in the 0.8 g sample.
Given that the sample contains 50% Fe2O3 by weight, we can find the amount of Fe2O3 as follows:
Amount of Fe2O3 = (50/100) x 0.8 g
Amount of Fe2O3 = 0.4 g
Step 2: Calculate the molar mass of Fe2O3.
The molar mass of Fe2O3 is calculated as follows:
Molar mass of Fe = 55.85 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of Fe2O3 = (2 x 55.85) + (3 x 16.00)
Molar mass of Fe2O3 = 159.70 g/mol
Step 3: Calculate the moles of Fe2O3.
Using the formula moles = mass / molar mass, we can calculate the moles of Fe2O3 as follows:
moles of Fe2O3 = 0.4 g / 159.70 g/mol
moles of Fe2O3 ≈ 0.00251 mol
Step 4: Calculate the moles of NH3 required for precipitation.
Based on the balanced equation for the reaction between NH3 and Fe(OH)3:
Fe2O3 + 6NH3 → 2Fe(OH)3 + 3H2O
We can see that 6 moles of NH3 are required for 1 mole of Fe2O3. Therefore, the moles of NH3 required can be calculated as:
moles of NH3 = 6 x moles of Fe2O3
moles of NH3 ≈ 6 x 0.00251 mol
moles of NH3 ≈ 0.01505 mol
Step 5: Calculate the required mass of NH3.
Knowing the mass of NH3 and its density, we can calculate its volume as follows:
Volume of NH3 = mass of NH3 / density of NH3
Volume of NH3 = (0.01505 mol x 17.03 g/mol) / 0.986 g/mL
Volume of NH3 ≈ 0.259 mL
Therefore, approximately 0.259 milliliters of NH3(aq) solution are required to precipitate iron as Fe(OH)3 in the 0.8 g sample.
To calculate the number of milliliters of NH3(aq) solution required, we need to follow a few steps:
Step 1: Calculate the mass of Fe2O3 in the 0.8 g sample.
- The sample is 50% Fe2O3, so the mass of Fe2O3 can be calculated as:
Mass of Fe2O3 = 0.8 g * (50/100) = 0.4 g
Step 2: Calculate the number of moles of Fe2O3.
- We need to convert the mass of Fe2O3 to moles using its molar mass. The molar mass of Fe2O3 can be calculated as:
Molar mass of Fe2O3 = (2 * atomic mass of Fe) + (3 * atomic mass of O) = (2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
- Now, we can calculate the number of moles of Fe2O3:
Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3 = 0.4 g / 159.69 g/mol
Step 3: Calculate the number of moles of NH3 required for precipitation.
- The reaction between Fe2O3 and NH3 is as follows:
Fe2O3 + 6NH3 -> 2Fe(OH)3 + 3(NH4)NO3
- From the balanced equation, we can see that 1 mole of Fe2O3 requires 6 moles of NH3 for precipitation.
Step 4: Calculate the mass of NH3 required for precipitation.
- Using the mole ratio from the balanced equation, we can calculate the number of moles of NH3 required:
Moles of NH3 = Moles of Fe2O3 * (6 moles of NH3 / 1 mole of Fe2O3)
- Now, calculate the mass of NH3 using its molar mass (17.03 g/mol):
Mass of NH3 = Moles of NH3 * Molar mass of NH3 = (Moles of Fe2O3 * 6) * 17.03 g/mol
Step 5: Calculate the volume of NH3(aq) solution using its density.
- We have the mass of NH3, and the density of the solution is given as 0.986 g/mL.
Volume of NH3(aq) = Mass of NH3 / Density of NH3(aq) = Mass of NH3 / 0.986 g/mL
By following these steps, you should be able to calculate the number of milliliters of NH3(aq) solution required to precipitate iron as Fe(OH)3 in the given sample.