If 10.0 L of hydrogen collected over water at 20' C and 750 mm Hg are produced, how many grams of iron metal must react with excess steam?
Do you have a reaction equation? One source gives Fe2O3 as the product; another gives Fe3O4 as the product. The answer to this question depends upon which product is formed. I will assume Fe2O3 and go from there.
2Fe + 3H2O ==> Fe2O3 + 3H2
Use PV = nRT and solve for n = mols H2 at the conditions listed.Remember T must be in kelvin and P in atmospheres (750 mm/760mm = ? atm)
Then convert mols H2 to mols Fe.
mols Fe needed = mols H2 x (2 mols Fe/3 mols H2) = mols H2 x 2/3 = ?
Finally, convert mols Fe to grams. g Fe = mols Fe x atomic mass Fe = ?
To determine the grams of iron metal required to react with excess steam, we need to generate the balanced chemical equation for the reaction and use stoichiometry.
The balanced chemical equation for the reaction between iron metal and steam (water vapor) is as follows:
3 Fe + 4 H2O -> Fe3O4 + 4 H2
From the balanced equation, we can see that 3 moles of iron (Fe) react with 4 moles of water (H2O) to produce 1 mole of iron(III) oxide (Fe3O4) and 4 moles of hydrogen gas (H2).
Now, let's convert the given volume of hydrogen gas collected over water to moles using the ideal gas law:
PV = nRT
Where:
P = Pressure (in atm)
V = Volume (in L)
n = Moles of gas
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (in K)
First, let's convert the pressure from mmHg to atm. We divide the pressure value by 760 mmHg/atm:
750 mmHg / 760 mmHg/atm ≈ 0.987 atm
Next, convert the given temperature in Celsius to Kelvin:
20°C + 273.15 = 293.15 K
Now, we can calculate the number of moles of hydrogen gas (H2):
n = (PV) / (RT)
n = (0.987 atm * 10.0 L) / (0.0821 L.atm/mol.K * 293.15 K)
n ≈ 0.4117 moles
According to the balanced equation, 3 moles of iron (Fe) react with 4 moles of water (H2O) to produce hydrogen gas (H2). Therefore, for every 4 moles of hydrogen gas produced, we need 3 moles of iron.
Now, we can calculate the moles of iron(Fe) required:
mol Fe = (0.4117 moles of H2) * (3 moles of Fe / 4 moles of H2)
mol Fe ≈ 0.3088 moles
Finally, we can calculate the mass of iron metal (Fe) using its molar mass:
molar mass of Fe = 55.845 g/mol
mass of Fe = (0.3088 moles of Fe) * (55.845 g/mol)
mass of Fe ≈ 17.229 g
Therefore, approximately 17.229 grams of iron metal must react with excess steam.