7.55 grams of P 4 and 7.55 grams of O 2 react according to the following reactio

Question

7.55 grams of P4 and 7.55 grams of O2 react according to the following reaction:

P4 + O2 --> P4O6

If enough oxygen is available, then the P4O6 reacts further:

P4O6 + O2 --> P4O10

a. Find the limiting reagent in the formation of P4O10.

b. What mass of P4O10 is produced?

c. What mass of excess reactant remains?

NOTE: This is a sequential action and limiting reagent problem. I asked this previously, and received answers that I thought were correct - but were not.

Make sure to show work, show units, and walk me through the problem to arrive at the correct answer. Make sure the work is correct as well, please. Thanks so much!

Explanation / Answer

a

P4(s) + 3O2(g) ? P4O6(s) and P4O6(s) + 2O2 ? P4O10

P4(s) + 5O2(g) ? P4O10(s)

Mols O2 = 0.180
Mols P4 = 0.0465

Mols O2/ Mols P4 - 3.87:1 Ratio
Needs to be 5:1, therefore, oxygen is limiting.

b

1. Balance the chemical equation.
4PH3(g) + 8O2(g) --> P4O10(s) + 6H2O(g)
Notice that 4 moles of PH3 produce 1 mole of P4O10.

2. Find the number of moles of PH3 in 225g.
Find the molecular mass of PH3
P= 31.0 to 3 significant figures
H= 1.0
PH3 = 31.0 + 3(1.0) = 34.0 g/mol
mol PH3 = 225g x 1 mol PH3/34.0 g = 6.62 mol PH3

3. Find number of moles of P4O10 using the coefficients from the balanced equation.
mol P4O10 = 6.62 mol PH3 x 1 mol P4O10/4 mol PH3 = 1.66 mol P4O10

4. Find number of grams of P4O10
Find the molar mass of P4O10
P: 4(31.0) = 124
O: 10(16.0) = 160
Molar mass of P4O10 = 284g/mol
gP4O10 = 1.66 mol P4O10 x 284g/1mol = 471 g P4O10

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