how many grams of Fe2O3 are formed when 16.7g of Fe reacts completely with oxygen.
4Fe +3O2 --> 2Fe2O3
how do i do this problem..thanks:)
To solve this problem, you will need to use stoichiometry. Stoichiometry is a method used to determine the quantities of reactants and products involved in a chemical reaction.
First, write down the balanced chemical equation for the reaction:
4Fe + 3O2 --> 2Fe2O3
From the equation, you can see that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3. This is the stoichiometric ratio.
Next, you need to calculate the number of moles of Fe present in the given mass. To do this, divide the given mass of Fe by its molar mass:
Molar mass of Fe = 55.845 g/mol
Moles of Fe = mass of Fe / molar mass of Fe
Moles of Fe = 16.7 g / 55.845 g/mol
Calculate the moles of Fe, which comes out to be approximately 0.2993 moles.
Now, using the stoichiometric ratio from the balanced chemical equation, you can calculate the moles of Fe2O3 formed. Since the stoichiometric ratio is 4:2 for Fe:Fe2O3, you divide the moles of Fe by 4 and multiply by 2 to get the moles of Fe2O3:
Moles of Fe2O3 = (Moles of Fe / 4) * 2
Moles of Fe2O3 = (0.2993 moles / 4) * 2
Calculate the moles of Fe2O3, which comes out to be approximately 0.1496 moles.
Finally, you need to convert the moles of Fe2O3 to grams by multiplying it by the molar mass of Fe2O3:
Molar mass of Fe2O3 = 159.6882 g/mol
Mass of Fe2O3 = Moles of Fe2O3 * Molar mass of Fe2O3
Mass of Fe2O3 = 0.1496 moles * 159.6882 g/mol
Calculate the mass of Fe2O3, which comes out to be approximately 23.87 grams.
Therefore, approximately 23.87 grams of Fe2O3 are formed when 16.7 grams of Fe reacts completely with oxygen.
To solve this problem, you need to use stoichiometry, which allows you to relate the amount of one substance in a balanced chemical equation to the amount of another substance. In this case, you want to determine the number of grams of Fe2O3 formed when 16.7g of Fe reacts completely with oxygen.
First, you need to calculate the molar mass of Fe (iron) and Fe2O3 (iron(III) oxide). The molar mass of Fe is 55.85 g/mol, and the molar mass of Fe2O3 is 159.69 g/mol. These values can be found on the periodic table.
Now, you can use the stoichiometric coefficients from the balanced equation to set up the conversion factor:
4 mol Fe = 2 mol Fe2O3
Using the molar masses, you can convert the grams of Fe to moles using the following conversion factor:
16.7 g Fe x (1 mol Fe / 55.85 g Fe) = 0.298 mol Fe
Now, you can use the stoichiometric conversion factor to find the moles of Fe2O3:
0.298 mol Fe x (2 mol Fe2O3 / 4 mol Fe) = 0.149 mol Fe2O3
Finally, you can convert moles of Fe2O3 to grams using the molar mass:
0.149 mol Fe2O3 x (159.69 g Fe2O3 / 1 mol Fe2O3) ≈ 23.8 g Fe2O3
Therefore, approximately 23.8 grams of Fe2O3 are formed when 16.7 grams of Fe reacts completely with oxygen.
(a) Look up the molar mass of Fe (= the atomic mass of Fe labeled "gFe/mol")
(b) Divide16.7gFe by (a) to get moles of Fe
(c) Multiply (b) by 2molFe2O3/4molFe (or 2/4) to get moles of Fe2O3 produced.
(d) multiply (c) by the formula mass of Fe2O3*
*To find the formula mass of Fe2O3, look up each atomic mass, multiply each by the subscript, and add add them all up.