over the years, the thermite reaction has been used for welding railorad rails, in incendiary bombs, and to ignite solid fuel rocket motors. The reaction:
Fe2O3(s) + 2 Al (s) -> 2 Fe(l) + Al2O3 (s)
What masses of iron ( III) oxide and aluminum must be used to produce 15.0 grams of iron? What is the maximum mass of aluminum oxide that could be produced?
We first check if the equation is balanced and according to our equation its already balanced.2nd we find the ratios from the equation.
FeO3+2AL=2Fe+AL2O3
2 = :2
x = :mass over
rmm=the answer you u will find when u do mol=mass/rmm.
Then n=mass/molar mass
Mass=mol x rmm
mass=0.134 x 159.6
21.359.final
To determine the masses of iron (III) oxide and aluminum needed to produce 15.0 grams of iron, we can use stoichiometry and the given balanced chemical equation.
Step 1: Calculate the molar mass of Fe2O3 (iron (III) oxide) and Al (aluminum).
- The molar mass of Fe2O3 = 2(55.845 g/mol) + 3(16.00 g/mol) = 159.70 g/mol
- The molar mass of Al = 26.98 g/mol
Step 2: Determine the number of moles of iron produced.
- In the balanced equation, the coefficient of Fe2O3 is 2, which means that 2 moles of Fe will be produced for every 1 mole of Fe2O3.
- Using the molar mass, we can convert grams of Fe to moles:
Moles of Fe = 15.0 g / (55.845 g/mol) = 0.2682 mol
Step 3: Determine the number of moles of Fe2O3 needed.
- Since the ratio of Fe2O3 to Fe is 1:2, we need to multiply the moles of Fe by 2 to get the moles of Fe2O3:
Moles of Fe2O3 = 0.2682 mol x 2 = 0.5364 mol
Step 4: Calculate the mass of Fe2O3 needed.
- Using the molar mass of Fe2O3, we can convert moles to grams:
Mass of Fe2O3 = 0.5364 mol x 159.70 g/mol = 85.62 g
Step 5: Calculate the mass of aluminum needed.
- In the balanced equation, the ratio of Al to Fe2O3 is 2:1.
- Therefore, the number of moles of Al needed is equal to half the number of moles of Fe2O3:
Moles of Al = 0.5364 mol / 2 = 0.2682 mol
- Using the molar mass of Al, we can convert moles to grams:
Mass of Al = 0.2682 mol x 26.98 g/mol = 7.23 g
Step 6: Calculate the maximum mass of aluminum oxide (Al2O3) that could be produced.
- Since the ratio of Al2O3 to Al is 1:2, the moles of Al2O3 will be equal to the moles of Al used.
- Using the molar mass of Al2O3, we can convert moles to grams:
Mass of Al2O3 = 0.2682 mol x (2 x 26.98 g/mol) = 14.55 g
Therefore, the mass of iron (III) oxide needed is 85.62 grams, the mass of aluminum needed is 7.23 grams, and the maximum mass of aluminum oxide that could be produced is 14.55 grams.
To find the masses of iron (III) oxide and aluminum needed to produce 15.0 grams of iron, we first need to determine the molar ratios between the reactants and products.
Looking at the balanced chemical equation:
Fe2O3(s) + 2 Al(s) -> 2 Fe(l) + Al2O3(s)
The stoichiometric ratio tells us that 1 mole of Fe2O3 reacts with 2 moles of Al to produce 2 moles of Fe and 1 mole of Al2O3.
1. Calculating the mass of Fe2O3 required:
First, we need to determine the molar mass of Fe2O3.
Fe: atomic mass = 55.85 g/mol
O: atomic mass = 16.00 g/mol (there are 3 oxygens in Fe2O3)
Molar mass of Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 159.70 g/mol
Now, we can set up a proportion to find the mass of Fe2O3 required:
(15.0 g Fe) / (1 mol Fe) = (x g Fe2O3) / (159.70 g Fe2O3)
Solving for x, we have:
x = (15.0 g Fe) * (159.70 g Fe2O3) / (1 mol Fe)
2. Calculating the mass of Al required:
The molar mass of Al is 26.98 g/mol.
Using the stoichiometric ratio, we know that for every 2 moles of Al, 1 mole of Fe2O3 is required. Therefore:
x = (15.0 g Fe) * (26.98 g Al) / (2 * 159.70 g Fe2O3)
Next, to find the maximum mass of aluminum oxide (Al2O3) that could be produced, we need to calculate the actual mass of Al2O3 formed and then subtract it from the initial mass of Al (which is determined above).
To calculate the mass of Al2O3:
Molar mass of Al2O3 = (2 * 26.98 g/mol) + (3 * 16.00 g/mol) = 101.96 g/mol
The stoichiometric ratio tells us that for every 2 moles of Al, 1 mole of Al2O3 is produced.
To find the maximum mass, we set up the same proportion:
x = (15.0 g Fe) * (101.96 g Al2O3) / (2 * 159.70 g Fe2O3)
Thus, the maximum mass of aluminum oxide that could be produced is x grams.
15g of Fe is 15/55.845 = 0.2686 moles
Each mole of Fe requires 1 mole of Fe2O3 and 2 moles of Al.
mol wt Fe2O3 = 2*55.845+3*15.999 = 159.687
mol wt Al = 26.982
so, we need
.2686*159.687 = 42.9g Fe2O3
.2686*2*26.982 = 14.5g Al
mol wt Al2O3 = 2*26.982+3*15.999 = 101.961
so, we get .2686*101.961 = 68.9g Al2O3