posted by tdb on .
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?
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.