Solid iron(III) hydroxide decomposes to produce
iron(III) oxide and water vapor. 1.09 L
of water vapor is produced at STP. How much
iron(III) hydroxide was used? in g.
Start by balancing the equation. Convert 1.09L H2O vapor to mols. Convert mols H2O to iron(III) hydroxide using the coefficients in the balanced equation, the convert that to grams. I can help you through it if you get stuck but explain in detail what you don't understand.
I have no idea!?!?
To find the amount of iron(III) hydroxide used, we can use the stoichiometry of the reaction and the molar volume of a gas at STP.
The balanced chemical equation for the reaction is:
2 Fe(OH)3 -> Fe2O3 + 3 H2O
From the equation, we can see that 2 moles of iron(III) hydroxide produce 3 moles of water vapor.
We know the volume of water vapor produced is 1.09 L at STP. However, we need to convert this volume to moles using the molar volume of a gas at STP, which is 22.4 L/mol.
1.09 L * (1 mol / 22.4 L) = 0.0487 mol
Now we can use the stoichiometry of the reaction to find the amount of iron(III) hydroxide:
2 moles Fe(OH)3 -> 3 moles H2O
x moles Fe(OH)3 -> 0.0487 moles H2O
x = (0.0487 mol * 2 mol Fe(OH)3) / 3 mol H2O
x = 0.0325 mol Fe(OH)3
Finally, we can convert the moles of iron(III) hydroxide to grams using its molar mass. The molar mass of Fe(OH)3 is:
1 mol Fe = 55.85 g
3 mol O = 3 * 16.00 g
3 mol H = 3 * 1.01 g
Molar mass of Fe(OH)3 = 55.85 g + 3 * 16.00 g + 3 * 1.01 g = 106.89 g/mol
0.0325 mol * 106.89 g/mol = 3.476 g
Therefore, approximately 3.476 grams of iron(III) hydroxide was used.
To determine the amount of iron(III) hydroxide used, you need to use the information given and utilize the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction:
Fe(OH)3(s) -> Fe2O3(s) + H2O(g)
From the equation, we can see that for every mole of Fe(OH)3, one mole of H2O is produced.
Next, we need to calculate the number of moles of H2O produced at STP. STP stands for standard temperature and pressure, which is 273 K (0 degrees Celsius) and 1 atm.
Using the ideal gas law, we can determine the number of moles of water vapor:
n = PV/RT
Where:
P = pressure (1 atm)
V = volume of water vapor (1.09 L)
R = ideal gas constant (0.0821 L·atm·mol^−1·K^−1)
T = temperature (273 K)
Plugging in the values:
n = (1 atm)(1.09 L) / (0.0821 L·atm·mol^−1·K^−1)(273 K)
n ≈ 0.0437 moles
Since the stoichiometric ratio between Fe(OH)3 and H2O is 1:1, this means that 0.0437 moles of Fe(OH)3 were used.
Finally, we can find the molar mass of Fe(OH)3, which is composed of iron (Fe), oxygen (O), and hydrogen (H):
Fe: 1 × atomic mass of Fe
O: 3 × atomic mass of O
H: 3 × atomic mass of H
Fe(OH)3: (1 × atomic mass of Fe) + (3 × atomic mass of O) + (3 × atomic mass of H)
Using the atomic masses:
Fe(OH)3: (1 × 55.85 g/mol) + (3 × 16.00 g/mol) + (3 × 1.01 g/mol)
Fe(OH)3: 55.85 g/mol + 48.00 g/mol + 3.03 g/mol
Fe(OH)3: 106.88 g/mol
Finally, multiplying the number of moles by the molar mass will give us the mass of Fe(OH)3 used:
Mass = n × molar mass
Mass = 0.0437 moles × 106.88 g/mol
Mass ≈ 4.68 grams
Therefore, approximately 4.68 grams of iron(III) hydroxide was used in the reaction.