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[rita_debbaneh2G]

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, P4O6: P4(s) 1 3 O2(g) S P4O6(s). If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus(V) oxide, P4O10: P4O6(s) 1 2 O2(g) S
P4O10(s). (a) What is the limiting reactant for the formation of P 4 O 10 ? (b) What mass of P 4 O 10 is produced? (c) How many grams of the excess reactant remain in the reaction vessel?

Can someone explain how they got the answer to this for me? None of the previous explanations are really helping.

[Olivia Smith 2E]

Calculate mols:
O2: 0.360625 mol
P4: 0.046532 mol

Put into equation and find ratio
original equation balanced:
P4 + 3O2 --> P4O6
Input values of mols
0.046532P4 + 0.360625/3O2 --> ??P4O6 We divide O2 mols by 3 to get a proper ratio of P4 to O2. For every one P4 3O2 is gonna be used so now we reduced it to a
New "Ratio" equation straight up one to one ratio
0.046532P4 + 0.1202083 O2 ---> 0.046532 P4O6 Take smallest name in the "ratio" equation and make that the mol value of the product this will also be your limiting reactant
We now have about 0.2210289 mol of O2 left; You may say wait that's way more than the "ratio" equation, that's right because the ratio equation is not the real amount that we have left, we just used to to figure out how much product we have. Make sure to convert it back to the actual amount at the end.

Next equation:
P4O6 + 2O2 --> P4O10
From the previous calculation we have
0.046532 P4O6 + 0.2210289/2 O2 --> ??P4O10 Do the same ratio thing with O2
Ratio Equation:
0.046532 P4O6 + .1105 O2 ---> 0.046532 P4O10 repeat the above steps, smallest value= product value= limiting reactant. In this case the limiting reactant will be the P4, or P4O6 if a part by part equation is assumed?
calculate molar mass of P4O10 160+ 124= 184g/mol
0.046532 x 184= 8.56g
There is 0.127936 mols of O2 left. Multiple by 16 =2.04 g