What number of moles of O2 is needed to produce 75 grams of P4O10 from P? (Molecular weight P4O10 = 284 g/mol)

Question

Answer choices

A.0.26 moles

B.2.64 moles

C 0.5 moles

D. 1.25 moles

Answer 1

Hello, Lela,

First, write a balanced equation for the reaction. That is:

4P + 5O2 = P4O10

(I wish they were all this easy)

Next, calculate the number of moles of the P4O10 are produced. The molecular weight of P4O10 is given as 284 g/mol. [I calculated 204 for this number, but since 284 was given, it might be in a hydrate form].

Assuming 284 g/mole is correct, the number of moles of P4O10 is calculated as:

(75g)/(284g.mole) = 0.2641 moles [the g cancles and moles moves to the top]. Rounded to 2 significant figures (since 75 is only 2 significant figures), the number of moles of P4O10 is 0.26.

The equations says we need 5 moles of O2 for every mole of P4O10. So this means we need:

5*0.26 moles, or 1.32 moles of O2, which should be reported as 1.3 moles (2 sig figs).

I hope this helps,

Bob

Answer 2

Hi Lela! Do me a favor and double check these numbers. I believe something is off.

We're using the chemical equation:

4P + 5O₂ → P₄O₁₀

First convert grams of product to moles:

75g P₄O₁₀ x (1mol / 284 g) = 0.264 mol P₄O₁₀

Using the mole ratios, convert moles of product into moles of reactant needed:

0.264 mol P₄O₁₀ x (5 mol P / 1 mol P₄O₁₀) = 1.32 mol O₂

See the same problem (with different starting g) here: http://msmorrischemistry.weebly.com/uploads/3/8/9/5/38951057/stoichiometry__ap_mc_.pdf