Oxygen gas reacts with powdered aluminum according to the following reaction:

4Al(s)+3O2(g)→2Al2O3(s)

What volume of O2 gas, measured at 787 mmHg and 21 ∘C, is required to completely react with 55.0 g of Al?
Express the volume in liters to three significant figures.

Considering the reaction stoichiometry and the ideal gas law, the volume of O₂ gas, measured at 787 mmHg and 21 ∘C, required to completely react with 55.0 g of Al is 35.47 L.

The balanced reaction is:

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Since the molar mass of Al is 27 g/mole, that is, the amount of mass that a substance contains in one mole, then the number of moles containing 55 grams of Al that react is calculated as:

Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 2.037 moles of Al will react with how many moles of O₂?

amount of moles of O₂= 1.53 moles

On the other side, an ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P× V = n× R× T

For O₂ gas you know:

Replacing in the ideal gas law:

1.04 atm× V = 1.53 moles× 0.082× 294 K

Solving:

V = (1.53 moles× 0.082× 294 K)÷ 1.04 atm

V= 35.47 L

Finally, the volume of O₂ gas, measured at 787 mmHg and 21 ∘C, required to completely react with 55.0 g of Al is 35.47 L.

Learn more:  

35.5L of O2

Explanation:

4Al(s) + 3O2(g) → 2Al2O3(s)

First, let us calculate the number of mole O2 that reacted with 55g of Al. This is illustrated below:

Molar Mass of Al = 27g/mol

Mass of Al = 55g

Number of mole = Mass /Molar Mass

Number of mole of Al = 55/27 = 2.04moles

From the equation,

4moles of Al reacted completely with 3 moles of O2.

Therefore, 2.04moles of Al will react completely with = (2.04 x 3)/4 = 1.53moles of O2

Now let us obtain the volume of O2 occupied by 1.53moles of O2 measured at 787 mmHg and 21°C. This can be achieved by doing the following:

n = 1.53moles

P = 787mmHg

Recall: 760mmHg = 1atm

787mmHg = 787/760 = 1.04atm

T = 21°C = 21 + 273 = 294K

R = 0.082atm.L/Kmol

V =?

Using the Ideal gas equation PV = nRT, we can obtain the volume as follows:

PV = nRT

V = nRT/P

V = (1.53 x 0.082 x 294)/1.04

V= 35.5L

Therefore, 35.5L of O2 required to react with 55.0g of Al

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole

it's c (engineer's do their work with mechanical stuff not on any living being)