Solud iron and steam react at high temperatures to form iron(vi) peroxide and hydrogen gas. How much solid iron is needed to obtained 405 grams of iron(vi) peroxide, with a 84.2% yield?
To determine how much solid iron is needed to obtain 405 grams of iron(VI) peroxide with an 84.2% yield, we need to follow these steps:
Step 1: Write out the balanced chemical equation.
The balanced chemical equation for the reaction between solid iron (Fe) and steam (H2O) to form iron(VI) peroxide (FeO3) and hydrogen gas (H2) is:
3Fe + 4H2O -> FeO3 + 4H2
Step 2: Calculate the molar mass of FeO3.
The molar mass of FeO3 is calculated by adding up the atomic masses of each element in the compound. The atomic mass of Fe is 55.85 g/mol, and the atomic mass of O is 16.00 g/mol. Since there are three oxygen atoms in FeO3, we can calculate the molar mass as follows:
Molar mass of FeO3 = (1 x 55.85) + (3 x 16.00) = 159.85 g/mol
Step 3: Calculate the theoretical yield of FeO3.
The theoretical yield is the maximum amount of FeO3 that can be obtained from the given amount of iron (Fe). It can be calculated by using the stoichiometry of the balanced equation.
From the balanced equation, we can see that for every 3 moles of Fe, we obtain 1 mole of FeO3.
Moles of FeO3 = Moles of Fe x (1 mole FeO3 / 3 moles Fe)
To find the moles of Fe, we need to divide the given mass of FeO3 by the molar mass of FeO3:
Moles of Fe = Mass of FeO3 / Molar mass of FeO3
Moles of Fe = 405 g / 159.85 g/mol
Step 4: Calculate the actual yield of FeO3.
The actual yield is the amount of FeO3 obtained in reality, which is given as 84.2% of the theoretical yield (based on the given percentage yield).
Actual yield of FeO3 = Theoretical yield of FeO3 x Percentage yield / 100
Actual yield of FeO3 = (Moles of Fe x (1 mole FeO3 / 3 moles Fe)) x 84.2 / 100
Step 5: Calculate the mass of solid iron needed.
Now we can calculate the mass of solid iron (Fe) needed to obtain the given amount of FeO3.
Mass of Fe = Moles of Fe x Molar mass of Fe
Using these steps, we can calculate the mass of solid iron needed to obtain 405 grams of iron(VI) peroxide with an 84.2% yield.