Textbook Problem M.11

[claire smith]

A reaction vessel contains 5.77 g of white phosphorus and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorus(III) oxide,
P4 + 3O2 yields P4O6. If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus(V) oxide, P4O6 +2O2 yields P4O10
. (a) What is the limiting reactant for the formation of P4O10? (b) What mass of
is produced? (c) How many grams of the excess reactant remain in the reaction vessel?

I am a little confused on how to determine part a) because it translates between the reactants of one equation and the products of the next.

[Christina Gigoux 1D]

Because it is asking for the limiting reactant for the formation of P4O10, the reactants would be P4O6 and O2. To find these you first need to determine the limiting reactant of the first equation and from there you can calculate how much of those reactants are consumed. With this you can find the amount of P4O6 produced. To find the amount of O2, you need to see how much of the O2 was consumed with the first reaction and subtract that amount from the original amount of O2 to determine how much is avaliable for the 2nd reaction.

[Rebekah Han 2K]

First, you're going to want to find the limiting reactant using the first equation. Once you determine the limiting reactant, use 5.77g of the limiting reactant to solve for how much of the excess reactant was used up. Then, you can subtract 5.77 g of the excess reactant with how many g was actually used up to find the grams of the excess reactant that was left over for the second equation.

The numbers you're going to want to use for the second equation come from how much P4O6 was produced in the first equation and the number of grams of excess reactant that was left over. Solve for g of P4O10 using the second equation, and you should be able to find the limiting reactant for the formation of P4O10.

Hope this helps!