Fe2O3 + 3H2 → 2Fe + 3H2O
Calculate the mass of iron produced when 450.0 grams of iron (III) oxide react with hydrogen gas.
If you can please show me the work so I can learn how to do it
2 Fe = 2 * 56 = 112 grams / 2 mols
3 O = 3 * 16 = 48 g / 3 mols
so
Fe2O3 = 160 g/mol
so
450 g * ( 1 mol / 160 g) = 2.81 mols of iron oxide
That will give me 2 * 2.81 = 5.63 mols of Fe
5.63 * 56 g/mol = 315 grams of Fe
The work by Anonymous is quite right and is presented very well. Here is a set of instructions that will work any stoichiometry problem.
Step 1. Write and balance the equation.
Step 2. Convert what you have into moles. moles = grams/molar mass OR if it is a solution then moles = molarity x liters or M x L = ?
Step 3. Using the coefficients, as Anonymous did above, convert moles of what you have to moles of what you want.
Step 4. If you want grams, then grams = moles x molar mass = ?
Good luck.
To calculate the mass of iron produced, we need to use the concept of stoichiometry. Here's how you can solve it step-by-step:
1. Determine the molar masses of the substances involved in the reaction:
- Fe2O3: (2 x molar mass of Fe) + (3 x molar mass of O)
- H2: 2 x molar mass of H
- Fe: molar mass of Fe
- H2O: 2 x molar mass of H + molar mass of O
The molar masses are:
- Fe2O3: (2 x 55.85 g/mol) + (3 x 16.00 g/mol) = 159.7 g/mol
- H2: 2 x 1.01 g/mol = 2.02 g/mol
- Fe: 55.85 g/mol
- H2O: (2 x 1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
2. Set up the balanced chemical equation:
Fe2O3 + 3H2 → 2Fe + 3H2O
3. From the balanced equation, we can determine the stoichiometric ratio between Fe2O3 and Fe. In this case, it's 2:2, which simplifies to 1:1.
4. Calculate the number of moles of Fe2O3:
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 450.0 g / 159.7 g/mol
≈ 2.82 mol
5. Use the stoichiometry to find the number of moles of Fe produced:
moles of Fe = moles of Fe2O3
6. Calculate the mass of Fe produced:
mass of Fe = moles of Fe × molar mass of Fe
= 2.82 mol × 55.85 g/mol
≈ 157.4 g
Therefore, approximately 157.4 grams of iron would be produced when 450.0 grams of iron (III) oxide reacts with hydrogen gas.
To calculate the mass of iron produced when 450.0 grams of iron (III) oxide react with hydrogen gas, we need to apply stoichiometry.
1. Start by writing the balanced chemical equation:
Fe2O3 + 3H2 → 2Fe + 3H2O
2. Determine the molar mass of Fe2O3:
Fe2O3: (2 x atomic mass of Fe) + (3 x atomic mass of O)
= (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
= 159.69 g/mol
3. Convert the given mass of Fe2O3 into moles:
moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 450.0 g / 159.69 g/mol
≈ 2.82 mol
4. Using the stoichiometric ratio from the balanced equation, determine the moles of Fe formed:
From the equation: 1 mol of Fe2O3 reacts to give 2 mol of Fe
moles of Fe = 2.82 mol of Fe2O3 x (2 mol of Fe / 1 mol of Fe2O3)
= 5.64 mol of Fe
5. Calculate the mass of Fe produced:
mass of Fe = moles of Fe x molar mass of Fe
= 5.64 mol x 55.845 g/mol
≈ 315.4 g
Therefore, the mass of iron produced when 450.0 grams of iron (III) oxide react with hydrogen gas is approximately 315.4 grams.