The reaction between solid sodium and iron(III) oxide is one in a series of reactions that inflates an automobile airbag.
6 Na(s) + Fe2O3(s) 3 Na2O(s) + 2 Fe(s)
If 171.0 g Na and 213.5 g Fe2O3 are used in this reaction, what is the mass of excess reactant that remains after the reaction is complete?
To find the excess reactant, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a chemical reaction, determining the maximum amount of product that can be formed.
We can find the limiting reactant by comparing the number of moles of each reactant and their stoichiometric coefficients in the balanced equation.
1. Convert the mass of each reactant to moles.
Moles of Na = Mass of Na / Molar mass of Na
Moles of Na = 171.0 g / 22.99 g/mol = 7.44 mol
Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
Moles of Fe2O3 = 213.5 g / (55.85 g/mol + (2 * 16.00 g/mol)) = 1.91 mol
2. Calculate the mole ratio between Na and Fe2O3 using the balanced equation.
From the balanced equation: 6 Na(s) + Fe2O3(s) → 3 Na2O(s) + 2 Fe(s)
6 moles of Na react with 1 mole of Fe2O3.
3. Calculate the number of moles required for the reaction using the mole ratio:
Moles of Fe2O3 required = Moles of Na / 6 = 7.44 mol / 6 = 1.24 mol
4. Compare the moles of Fe2O3 required with the moles of Fe2O3 actually present:
Since we have 1.91 mol of Fe2O3 and only need 1.24 mol, Fe2O3 is in excess.
5. Calculate the moles of excess Fe2O3:
Excess moles of Fe2O3 = Moles of Fe2O3 - Moles of Fe2O3 required
Excess moles of Fe2O3 = 1.91 mol - 1.24 mol = 0.67 mol
6. Convert the moles of excess Fe2O3 to grams:
Mass of excess Fe2O3 = Moles of Fe2O3 x Molar mass of Fe2O3
Mass of excess Fe2O3 = 0.67 mol x (55.85 g/mol + (2 * 16.00 g/mol)) = 80.23 g
Therefore, the mass of excess reactant (Fe2O3) that remains after the reaction is complete is 80.23 grams.
To find the mass of excess reactant that remains after the reaction is complete, we first need to determine the limiting reactant.
1. Calculate the number of moles for each reactant:
Number of moles of Na = mass / molar mass = 171.0 g / 22.99 g/mol = 7.44 mol
Number of moles of Fe2O3 = mass / molar mass = 213.5 g / 159.69 g/mol = 1.34 mol
2. Use the stoichiometry of the balanced equation to determine the stoichiometric ratio between the reactants:
For every 6 moles of Na, you need 1 mole of Fe2O3.
3. Calculate the moles of Fe2O3 required to react with all the Na:
Moles of Fe2O3 required = (6 moles of Na / 1 mole of Fe2O3) x (7.44 moles of Na) = 44.64 moles of Fe2O3
Now, let's compare the moles of Fe2O3 required to the moles of Fe2O3 available:
- If the moles of Fe2O3 available is greater than the moles required, then Na is the limiting reactant and Fe2O3 is in excess.
- If the moles of Fe2O3 available is less than the moles required, then Fe2O3 is the limiting reactant and Na is in excess.
4. Compare the moles to determine the limiting reactant:
Moles of Fe2O3 available = 1.34 mol
Moles of Fe2O3 required = 44.64 mol
Since 1.34 mol is less than 44.64 mol, Fe2O3 is the limiting reactant.
5. Calculate the moles of Na required to react with all the Fe2O3:
Moles of Na required = (1 mole of Fe2O3 / 6 moles of Na) x (44.64 moles of Fe2O3) = 7.439 moles of Na
6. Calculate the mass of excess reactant that remains:
Mass of excess Na = (moles of Na available - moles of Na required) x molar mass
Mass of excess Na = (7.44 moles - 7.439 moles) x 22.99 g/mol = 0.022 g
Therefore, the mass of excess reactant that remains after the reaction is complete is 0.022 grams.
1. Write the equation and balance it. You need to write in an arrow so you can tell where the reactants end and the products start.
2a. Convert 171.0 g Na to moles. moles = grams/molar mass.
2b. Do the same for 213.6 g Fe2O3.
3a. Using the coefficients in the balanced equation, convert moles Na to moles of ANY product. I would choose Fe.
3b. Do the same for converting moles Fe2O3 to moles of ANY product but choose the SAME one you chose in 3a.
3c. Likely you will have two different answers for the product; obviously, both can't be correct (unless they are the same). The correct answer is ALWAYS the smaller one and the reactant producing that smaller one is the limiting reagent. The other reagent, of course, is the excess reagent.
4. Using the coefficients in the balanced equation, convert moles of the limiting reagent (now that you have it identified) to moles of the excess reagent and convert that to grams. grams = moles x molar mass. That will be number of grams of excess reagent used.
5. You know the excess reagent. You know how many grams you had initially. Subtract the grams used to find the amount remaining.
Post your work if you get stuck.