When heated solid iron3 hydroxide decomposes to produce iron 3 oxide and water vapor if 0.75L of water vapor is produced at 227 degrees celsius and 1.0 aTm. how many grams of FE(OH)3 were used?
To calculate the number of grams of Fe(OH)3 used, we need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
We need to convert the given temperature from Celsius to Kelvin by adding 273.15:
T = 227°C + 273.15 = 500.15 K
Now, let's rearrange the ideal gas law equation to solve for n (number of moles):
n = PV / RT
The pressure is given as 1.0 atm, and the volume is given as 0.75 L of water vapor. Substituting these values and the gas constant into the equation, we can calculate the number of moles:
n = (1.0 atm) * (0.75 L) / (0.0821 L·atm/(mol·K) * 500.15 K)
Simplifying, we find:
n = 0.018 mol
From the balanced chemical equation, we can see that the stoichiometric ratio between FE(OH)3 and water vapor is 1:3. This means that for every 1 mole of Fe(OH)3, 3 moles of water vapor are produced.
Thus, the number of moles of Fe(OH)3 used is also 0.018 mol.
To find the mass of Fe(OH)3, we need to use its molar mass, which can be calculated by adding the atomic masses of its constituent elements:
Molar mass of Fe = 55.845 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.008 g/mol
Molar mass of Fe(OH)3 = (55.845 g/mol) + 3 * [(16.00 g/mol) + (1.008 g/mol)]
Simplifying:
Molar mass of Fe(OH)3 = 106.848 g/mol
Finally, we can find the mass of Fe(OH)3 used by multiplying the number of moles by the molar mass:
Mass = n * Molar mass
Mass = 0.018 mol * 106.848 g/mol
Calculating, we find:
Mass = 1.92 g
Therefore, approximately 1.92 grams of Fe(OH)3 were used.