Iron (iii) hydroxide reacts with acetic acid to form iron (iii) acetate and water. If 45.4 grams of water are formed, how many grams of iron (iii) acetate will be produced?
Fe(OH)3 + 3HC2H2O2 --> Fe(C2H3O2)3 + 3H2O
Fe(OH)3 + 3HC2H2O2 --> Fe(C2H3O2)3 + 3H2O
How many mols in 45.4 g H2O. mols = g/molar mass
= 45.4/18 = 2.5 but that's an estimate. You should redo all of these calculations. Now use the coefficients in the equation.
Then 2.5 mols H2O x ( 1 mol Fe(C2H3O2)3/3 mols H2O) = mols Fe(C2H3O2)3.
Convert that to grams with the above formula.
To calculate the grams of iron (III) acetate produced, we first need to find the molar masses of the compounds involved.
The molar mass of Fe(OH)3 (iron (III) hydroxide) is calculated by summing the molar masses of its constituents:
Fe: 55.85 g/mol
O: 16.00 g/mol (there are 3 oxygen atoms)
H: 1.01 g/mol (there are 3 hydrogens)
Fe(OH)3:
55.85 g/mol + (16.00 g/mol x 3) + (1.01 g/mol x 3) = 106.86 g/mol
The molar mass of H2O (water) is calculated by summing the molar masses of its constituents:
H: 1.01 g/mol (there are 2 hydrogens)
O: 16.00 g/mol
H2O:
1.01 g/mol x 2 + 16.00 g/mol = 18.02 g/mol
Now, we can calculate the moles of water produced using its mass:
moles of H2O = mass / molar mass
moles of H2O = 45.4 g / 18.02 g/mol = 2.52 mol
From the balanced equation, we know that 1 mole of Fe(OH)3 reacts with 3 moles of H2O to produce 1 mole of Fe(C2H3O2)3 (iron (III) acetate).
Therefore, moles of Fe(C2H3O2)3 = moles of H2O.
moles of Fe(C2H3O2)3 = 2.52 mol
Finally, we can calculate the grams of iron (III) acetate produced using its molar mass:
grams of Fe(C2H3O2)3 = moles of Fe(C2H3O2)3 x molar mass of Fe(C2H3O2)3
The molar mass of Fe(C2H3O2)3 (iron (III) acetate) can be calculated by summing the molar masses of its constituents:
Fe: 55.85 g/mol
C: 12.01 g/mol (there are 6 carbon atoms)
H: 1.01 g/mol (there are 9 hydrogens)
O: 16.00 g/mol (there are 6 oxygen atoms)
Fe(C2H3O2)3:
55.85 g/mol + (12.01 g/mol x 2) + (1.01 g/mol x 3) + (16.00 g/mol x 3) = 232.99 g/mol
Therefore, grams of Fe(C2H3O2)3 = 2.52 mol x 232.99 g/mol = 587.48 grams.
So, 587.48 grams of iron (III) acetate will be produced.
To determine the grams of iron (III) acetate produced, we need to calculate the molar mass of water (H2O) and iron (III) acetate (Fe(C2H3O2)3). Then, we can use stoichiometry to find the grams of iron (III) acetate from the given mass of water.
The molar mass of water (H2O) can be calculated as:
H: 1.01 grams/mol * 2 = 2.02 grams/mol
O: 16.00 grams/mol * 1 = 16.00 grams/mol
Total molar mass of water (H2O) is 18.02 grams/mol.
Next, let's calculate the molar mass of iron (III) acetate (Fe(C2H3O2)3):
Fe: 55.85 grams/mol * 1 = 55.85 grams/mol
C: 12.01 grams/mol * 2 = 24.02 grams/mol
H: 1.01 grams/mol * 3 * 2 = 6.06 grams/mol
O: 16.00 grams/mol * 2 * 3 = 96.00 grams/mol
Total molar mass of iron (III) acetate (Fe(C2H3O2)3) is 181.93 grams/mol.
Now, we can use stoichiometry to find the grams of iron (III) acetate produced from the given mass of water:
1 mole of Fe(OH)3 reacts with 3 moles of H2O to form 1 mole of Fe(C2H3O2)3.
The number of moles of water (H2O) produced can be calculated as:
45.4 grams * (1 mol/18.02 grams) = 2.52 moles
Since the ratio of moles between water (H2O) and iron (III) acetate (Fe(C2H3O2)3) is 3:1, we can use this ratio to find the number of moles of iron (III) acetate:
2.52 moles H2O * (1 mol Fe(C2H3O2)3/3 mol H2O) = 0.84 moles Fe(C2H3O2)3
Finally, to find the grams of iron (III) acetate produced, multiply the number of moles by the molar mass:
0.84 moles * 181.93 grams/mol = 152.87 grams
Therefore, approximately 152.87 grams of iron (III) acetate will be produced.