Question

How many grams of P4O10 (292.88 g/mol) form when phelpsphorous (P4, 125.52 g/mol) reacts with 16.2 L of O2 (33.472 g/mol) ) at standard temperature and pressure

Answer 1

After determining the moles of oxygen gas at STP, we use the balanced chemical reaction to find the moles of P4O10 produced. With the molar mass of P4O10, we then calculate the mass of P4O10 formed, which is 42.364 g.

Step-by-step explanation:

To calculate how many grams of P4O10 form when phosphorous reacts with O2, we first need to find the moles of O2. At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 L, so the moles of O2 are:

Moles of O2 = 16.2 L / 22.4 L/mol = 0.72321 mol O2

According to the balanced chemical equation:

P4(s) + 5O2(g) → P4O10(s)

1 mole of P4 reacts with 5 moles of O2 to produce 1 mole of P4O10. So the amount of P4O10 formed is limited by the initial amount of O2, which in this case is 0.72321 mol. Since the ratio is 5:1 for O2 to P4O10, the amount of P4O10 formed from the O2 is:

Moles of P4O10 = 0.72321 mol O2 / 5 = 0.14464 mol P4O10

Finally, to find the mass:

Mass of P4O10 = 0.14464 mol P4O10 × 292.88 g/mol = 42.364 g P4O10

Answer 2

40.5 g of P₄O₁₀ are produced

Step-by-step explanation:

We state the reaction:

P₄ + 5O₂ → P₄O₁₀

We do not have data from P₄ so we assume, it's the excess reactant.

We need to determine mass of oxygen and we only have volumne so we need to apply density.

Density = mass / volume, so Mass = density . volume

Denstiy of oxygen at STP is: 1.429 g/L

1.429 g/L . 16.2L = 23.15 g

We determine the moles: 23.15 g . 1mol / 33.472g = 0.692 moles

5 moles of O₂ can produce 1 mol of P₄O₁₀

Our 0.692 moles may produce (0.692 . 1)/ 5 = 0.138 moles

We determine the mass of product:

0.138 mol . 292.88 g/mol = 40.5 g