2Fe+3H2O→Fe2O3+3H2
If 1.24 mol of water react, how many g of iron(III) oxide would be produced?
I know that the answer to this is 66.0 g but I don't understand how to solve it.
It takes 3 moles of H2O to produce 1 mole of Fe2O3
so, since you have 1.24 moles of H20, you will get 0.413 moles of Fe2O3.
So, how many grams is that?
Fe2O3 has 2 atoms of Fe (55.845 g/mol)
and 3 atoms of O (16g/mol)
so, a mole of Fe2O3 is 2*55.845 + 3*16 = 159.69 grams
You have only 0.413 moles = 0.413 * 159.69 = 65.95 grams
or 66.0 if you round to the nearest 0.1 gram.
I'm sorry, I don't understand the next step.
mols = grams/molar mass. You have mols and you can calculate molar mass; substitute and solve for grams.
To solve this problem, you can use the balanced chemical equation provided:
2Fe + 3H2O → Fe2O3 + 3H2
The balanced equation indicates that 2 moles of Fe reacts with 3 moles of H2O to produce 1 mole of Fe2O3.
First, you need to determine the number of moles of H2O that react. The problem states that 1.24 mol of H2O is present.
Using the stoichiometric coefficients from the balanced equation, you can set up a ratio:
1 mole of H2O reacts with 1 mole of Fe2O3
So, if 3 moles of H2O react, you can set up a proportion:
3 moles H2O / 2 moles Fe = 1.24 moles H2O / x moles Fe2O3
Cross-multiplying gives:
3 * x = 2 * 1.24
Simplifying:
3x = 2.48
Dividing both sides by 3:
x = 0.8267 mol Fe2O3
Now, you can convert the moles of Fe2O3 to grams using the molar mass of Fe2O3. The molar mass of Fe2O3 can be calculated by adding up the atomic masses of its constituent elements:
Fe = 55.85 g/mol
O = 16.00 g/mol (there are three oxygen atoms, so multiply by 3)
Molar mass of Fe2O3:
2 * 55.85 g/mol + 3 * 16.00 g/mol = 159.69 g/mol
Finally, you can convert moles to grams:
0.8267 mol Fe2O3 * 159.69 g/mol = 131.99 g
Rounding to the appropriate number of significant figures, the answer is approximately 132 g of iron(III) oxide.