Can anyone please help me with this problem?? =/

The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflates an automobile airbag. If 100g of sodium and 100g of iron(III) oxide are used in this reaction, determine:

a. the balanced reaction
b. the limiting reaction
c. the reactant in excess
d. mass of the solid iron produced
e. mass of excess reactant that remains.

THANK YOU :)

Sure, I can help you with this problem! Let's go step by step:

a. The balanced reaction:
To determine the balanced reaction, we need to write out the chemical equation and balance the number of atoms on each side. The reaction is between solid sodium (Na) and iron(III) oxide (Fe2O3). The balanced reaction is as follows:

4 Na + Fe2O3 -> 2 Na2O + 2 Fe

b. The limiting reactant:
The limiting reactant is the one that gets completely consumed in the reaction and determines the amount of product formed. To find the limiting reactant, we need to calculate the amount of product that can be formed using each reactant. We can compare the moles of each reactant to the stoichiometric ratio in the balanced equation.

First, let's convert the masses of sodium and iron(III) oxide to moles. The molar mass of Na is 22.99 g/mol, and the molar mass of Fe2O3 is 159.69 g/mol.

For sodium:
Number of moles of Na = mass of Na / molar mass of Na
= 100 g / 22.99 g/mol

For iron(III) oxide:
Number of moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 100 g / 159.69 g/mol

Next, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. According to the balanced equation, it takes 4 moles of Na to react with 1 mole of Fe2O3.

Therefore:
Moles of Na required = moles of Fe2O3 * 4
= (100 g / 159.69 g/mol) * 4

If the number of moles of Na is greater than the calculated value, then Fe2O3 is the limiting reactant. If the number of moles of Na is less than the calculated value, then Na is the limiting reactant.

b. The reactant in excess:
The reactant that is not the limiting reactant is considered the excess reactant. So the reactant in excess can be determined based on the result of the previous step.

c. The mass of the solid iron produced:
Since the limiting reactant determines the amount of product formed, we need to determine the moles of the limiting reactant that are consumed and then calculate the mass of solid iron produced.

According to the balanced equation, 1 mole of Fe2O3 produces 2 moles of Fe. So the number of moles of Fe produced is equal to half the number of moles of Fe2O3 consumed.

Mass of solid iron produced = moles of Fe2O3 consumed * molar mass of Fe
= (moles of Fe2O3 * 1/2) * molar mass of Fe

d. The mass of excess reactant that remains:
To find the mass of excess reactant that remains, we need to determine the moles of the excess reactant that are not consumed and then calculate the mass.

Moles of excess reactant remain = Moles of excess reactant initially - Moles of excess reactant consumed

Then, the mass of excess reactant that remains can be calculated as:

Mass of excess reactant that remains = Moles of excess reactant remain * Molar mass of the excess reactant

I hope this helps! Let me know if you have any further questions.

I must tell you that I am unfamiliar with this in an air bag system; however, the only reaction that makes sense to me with these reactants is as follows:

6Na + Fe2O3 ==> 3Na2O + 2Fe
mols Na = g/atomic mass = ?
mols Fe2O3 = g/molar mass = ?
Convert mols Na to mols Fe solid using the coefficients in the balanced equation.
Do the same for mols Fe2O3 converted to mols Fe.
b.
It is likely that the two numbers for mols Fe will not agree; the correct answer in limiting reagent problems is ALWAYS the smaller one and the reactant producing that value is the limiting reagent.
c.
Of course the other reactant is the one in excess.
d.mols Fe x atomic mass = grams Fe.

e. Convert mols of the limiting reagent to mols of the excess reagent, multiply mols of the excess reagent x molar mass t obtain grams and subtract this number from g excess reagent there initially to find excess reagent remaining un-reacted.