Limiting Reactant Question

[Jaime_Chamberlain_3G]

Can you please explain how to do M.11 on the review questions? A reaction vessel contains 5.77g of white phosphorus and 5.77g of oxygen. The first reaction to take place is the formation of phosphorus(III) oxide, P4)6: P4(s) + 3O2(g)--> P4O6(s). If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus(V) oxide, P4O10: P4O6(s) + 2O2---> P4O10(s). a) What is the limiting reactant for the formation of P4O10? B) What mass of P4O10 is produced? C) How many grams of the excess reactant remain in the reaction vessel?

[Alexander_Chang_1I]

Basically you first have to determine which product is the limiting reactant by converting both O and P to moles and comparing the ratio (which is 4 Oxygen:10 Phosphorous), use the limiting reactant to determine the moles of P4O6 created in the first reaction, and use that to determine the moles of P4O10 created in the second reaction, which can be easily converted back to grams. From there you can work backwards to determine how many moles of P were used, convert to grams, and subtract the difference from the original 5.77g of P.

[Hao 1I]

A) To find the limiting reagant for the formation of P4O10, you would use stoichiometry to compare the usage of oxygen by P4 in both reactions. Given that 5.77g of oxygen is in the vessel, determine if there is enough oxygen present for the following reaction.
B) Knowing that you have 5.77g of oxygen present, determine the excess amount available for the production of P4O10
C) Knowing that you have 5.77g of P4 present, you would use stoich to determine the amount of P4O6 produced and compare it to the amount of actual amount that would have been produced by the limiting reagant from part a.
Hope this helps! - Hao