During a lab experiment 44.18 grams of al2o3 was formed with o2 reacted with aluminum metal at 290 k and 1.3 atm. what was the volume of o2 used during the experiment

11.9 L

Explanation:

Data Given:

Mass of Al₂O₃ = 44.18 g

Temperature = 290 K

Pressure = 1.3 atm

Volume = ?

Solution:

As aluminum metal react with Oxygen, therefore first look for the reaction

                4Al + 3O₂ > 2Al₂O₃

So from here we will get information about moles of oxygen

First find number of moles of  44.18 grams of Al₂O₃

Molar mass of Al₂O₃ = 102 g/mol

convert mass to moles

                    no.of moles = mass in g / molar mass

put values in above formula

                    no.of moles = 44.18 g / 102 g/mol

                    no.of moles = 0.433 mol

Now as we have the following equation

                              4Al  +    3O₂  >  2Al₂O₃

                                          3 mole                2 mole

So according to equation 2 mole Al₂O₃ form from 3 moles of oxygen the how many moles of oxygen will form 0.433 mole of Al₂O₃

                                3 moles of O₂  ≅  2 moles of Al₂O₃

                                X moles of O₂  ≅  0.433 moles of Al₂O₃

By cross multiplication

                                X moles of O₂ = 3 x 0.433 /2

                               moles of oxygen = 0.65 moles

Now find the Volume of Oxygen gas

for this formula used

                         PV = nRT

                           V = nRT / P. . . . . . . . . .  (1)

where  

n = no. of moles of O₂ = 0.65 mol

R = Gas Constant = 0.0821 L.atm.mol⁻¹.K⁻¹

Put values above formula (1)

               V = 0.65 mol x 0.0821 L.atm.mol⁻¹.K⁻¹ x 290 K /  1.3 atm.

               V = 11.9 L

So the volume of oxygen during experiment = 11.9 L

14 days is the half-life of phosphorus-32