Iron reacts with superheated steam(H2O) to form hydrogen gas and the oxide Fe2O3. (write) Balance this chemical reaction. And if 10grams of Iron react with excess steam, how many grams of Fe2O3 are produced? How many grams of H gas? How many grams of H2O consumed???
Here is the balanced equation.
4Fe + 6H2O ==> 6H2 + 2Fe2O3
Here is a link to a worked example of a stoichiometry problem. Just follow the steps to work any stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
The balanced chemical reaction for the reaction between iron (Fe) and superheated steam (H2O) to form hydrogen gas (H2) and iron(III) oxide (Fe2O3) can be written as:
3 Fe + 4 H2O -> Fe2O3 + 4 H2
To determine the grams of Fe2O3 produced, we need to calculate the molar mass of Fe2O3:
Molar mass of Fe2O3 = (2 * Molar mass of Fe) + (3 * Molar mass of O)
= (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
= 111.69 g/mol + 48.00 g/mol
= 159.69 g/mol
To calculate the grams of Fe2O3 produced from 10 grams of iron, we can use stoichiometry and the molar ratio from the balanced equation:
10 g Fe * (1 mol Fe / 55.845 g Fe) * (1 mol Fe2O3 / 3 mol Fe) * (159.69 g Fe2O3 / 1 mol Fe2O3) = 95.43 g Fe2O3
Therefore, 10 grams of iron will produce 95.43 grams of Fe2O3.
To calculate the grams of hydrogen gas (H2) produced, we can again use stoichiometry and the molar ratio from the balanced equation:
10 g Fe * (1 mol Fe / 55.845 g Fe) * (4 mol H2 / 3 mol Fe) * (2.02 g H2 / 1 mol H2) = 7.24 g H2
Therefore, 10 grams of iron will produce 7.24 grams of H2.
As for the grams of water (H2O) consumed, we can use the given information that there is an excess of steam. This means that all the water will be consumed in the reaction. Therefore, the grams of H2O consumed will be equal to the initial amount of water, which is not specified in the question.
To balance the chemical reaction, we can follow these steps:
Step 1: Write the unbalanced equation:
Iron (Fe) + Steam (H2O) → Hydrogen (H2) + Iron(III) oxide (Fe2O3)
Step 2: Balance the equation by adjusting the coefficients in front of each compound. Start by balancing the least abundant atoms, which in this case are the hydrogen atoms and the oxygen atoms.
Fe + H2O → H2 + Fe2O3
Next, balance the oxygen atoms. There are 3 oxygen atoms in Fe2O3 and 1 oxygen atom in H2O. To balance them, we need to add a coefficient of 3 in front of H2O.
Fe + 3H2O → H2 + Fe2O3
Finally, balance the hydrogen atoms by adding a coefficient of 4 in front of the H2.
Fe + 3H2O → 4H2 + Fe2O3
Now the equation is balanced.
To determine the amount of Fe2O3 produced when 10 grams of Iron reacts with excess steam, we need to use stoichiometry.
Step 1: Write and balance the equation:
Fe + 3H2O → 4H2 + Fe2O3
Step 2: Calculate the molar mass of Fe:
Fe = 55.85 g/mol
Step 3: Convert grams of Fe to moles using the molar mass:
10 g Fe * (1 mol Fe / 55.85 g Fe) = 0.179 mol Fe
Step 4: Use the mole ratio between Fe and Fe2O3 from the balanced equation (1:1) to find the moles of Fe2O3:
0.179 mol Fe * (1 mol Fe2O3 / 1 mol Fe) = 0.179 mol Fe2O3
Step 5: Convert moles of Fe2O3 to grams using the molar mass:
0.179 mol Fe2O3 * (159.69 g Fe2O3 / 1 mol Fe2O3) = 28.55 g Fe2O3
Therefore, when 10 grams of Iron react with excess steam, 28.55 grams of Fe2O3 are produced.
To find the amount of hydrogen gas (H2) and water (H2O) consumed, we can use the same steps.
For H2:
Step 1: Use the balanced equation to determine the mole ratio between Fe and H2 (1:4).
Step 2: Calculate the moles of H2 produced using the moles of Fe.
Step 3: Convert moles of H2 to grams.
For H2O:
Step 1: Use the balanced equation to determine the mole ratio between Fe and H2O (1:3).
Step 2: Calculate the moles of H2O consumed using the moles of Fe.
Step 3: Convert moles of H2O to grams.
Please provide the exact mass of Hydrogen gas(H2) and water consumed in grams so we can calculate it for you.