how many grams of iron (iii) nitrate are needed to produce 25.0 g of iron (iii) sulfate form sulfuric acid?
To determine the number of grams of iron (III) nitrate needed to produce 25.0 g of iron (III) sulfate from sulfuric acid, we need to follow these steps:
Step 1: Write and balance the chemical equation.
The balanced equation for the reaction between iron (III) nitrate and sulfuric acid to produce iron (III) sulfate is:
2 Fe(NO3)3 + 3 H2SO4 -> Fe2(SO4)3 + 6 HNO3
Step 2: Determine the molar mass of iron (III) nitrate.
The molar mass of iron (III) nitrate (Fe(NO3)3) can be calculated by summing up the molar masses of its constituent elements. Iron (Fe) has a molar mass of 55.85 g/mol, nitrogen (N) has a molar mass of 14.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. As there are three nitrate ions (NO3-) in the compound, we need to add the molar mass of one nitrogen atom and three times the molar mass of an oxygen atom:
Molar mass of Fe(NO3)3 = (1 * 55.85) + (1 * 14.01) + (3 * (1 * 16.00)) = 241.86 g/mol
Step 3: Apply stoichiometry.
From the balanced equation, we can see that 2 moles of iron (III) nitrate (Fe(NO3)3) react to produce 1 mole of iron (III) sulfate (Fe2(SO4)3). Therefore, the molar ratio is 2:1.
Step 4: Calculate the number of moles of iron (III) sulfate.
We need to convert the given mass of iron (III) sulfate to moles. The molar mass of iron (III) sulfate (Fe2(SO4)3) can be calculated by summing up the molar masses of its constituent elements. Iron (Fe) has a molar mass of 55.85 g/mol, sulfur (S) has a molar mass of 32.06 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol:
Molar mass of Fe2(SO4)3 = (2 * 55.85) + (3 * 32.06) + (12 * 16.00) = 399.88 g/mol
Number of moles of Fe2(SO4)3 = Mass / Molar mass = 25.0 g / 399.88 g/mol
Step 5: Calculate the number of moles of iron (III) nitrate.
Using the molar ratio obtained from the balanced equation, we can calculate the number of moles of iron (III) nitrate.
Number of moles of Fe(NO3)3 = Number of moles of Fe2(SO4)3 * (2 moles Fe(NO3)3 / 1 mole Fe2(SO4)3)
Step 6: Calculate the mass of iron (III) nitrate.
Finally, we can calculate the mass of iron (III) nitrate by multiplying the number of moles obtained in Step 5 with its molar mass.
Mass of Fe(NO3)3 = Number of moles of Fe(NO3)3 * Molar mass of Fe(NO3)3
By following these steps and performing the calculations, you will be able to determine the mass of iron (III) nitrate needed to produce 25.0 g of iron (III) sulfate from sulfuric acid.
To determine the number of grams of iron (III) nitrate needed to produce iron (III) sulfate from sulfuric acid, we first need to write and balance the chemical equations for the reactions involved.
The balanced equation for the reaction between iron (III) nitrate and sulfuric acid is:
Fe(NO3)3 + 3H2SO4 → Fe2(SO4)3 + 6HNO3
From the equation, we can see that one mole of iron (III) nitrate reacts with three moles of sulfuric acid to produce one mole of iron (III) sulfate. Therefore, we can use the molar mass to calculate the number of moles of iron (III) nitrate needed.
The molar mass of iron (III) nitrate (Fe(NO3)3) is:
1 iron (Fe) atom: 55.85 g/mol
3 nitrogen (N) atoms: 3 x 14.01 g/mol = 42.03 g/mol
9 oxygen (O) atoms: 9 x 16.00 g/mol = 144.00 g/mol
Total molar mass of Fe(NO3)3 = 55.85 + 42.03 + 144.00 = 241.88 g/mol
Now we can calculate the number of moles of iron (III) nitrate required:
Moles of Fe(NO3)3 = mass (g) / molar mass (g/mol)
Moles of Fe(NO3)3 = 25.0 g / 241.88 g/mol = 0.1033 mol
Since one mole of iron (III) nitrate reacts to produce one mole of iron (III) sulfate, we can say that 0.1033 mol of iron (III) nitrate is required.
Finally, we can calculate the mass of iron (III) nitrate needed:
Mass of Fe(NO3)3 = moles x molar mass
Mass of Fe(NO3)3 = 0.1033 mol x 241.88 g/mol = 25.0 g
Therefore, you would need 25.0 grams of iron (III) nitrate to produce 25.0 g of iron (III) sulfate from sulfuric acid.