Question

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s) + 3 O2(g) --> 2 Al2O3(s)0.5 x 0.67 x 22.42 x 1.5 x 22.40.5 x 1.5 x 22.42 x 0.67 x 22.4

Answer 1

34.28 L ( 1.5*22.4 L)

Step-by-step explanation:

Calculation of the moles of aluminum as:-

Mass = 55 g

Molar mass of aluminum = 26.981539 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (55\ g)/(26.981539\ g/mol)`

`Moles= 2.0384\ mol`

According to the reaction:-

`4Al+3O_2\rightarrow 2Al_2O_3`

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with
`(3)/(4)` moles of oxygen gas

2.0384 moles of aluminum react with
`(3)/(4)* 2.0384` moles of oxygen gas

Moles of oxygen gas = 1.5288 moles

At STP,

Pressure = 1 atm

Temperature = 273.15 K

Using ideal gas equation as:

`PV=nRT`

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1 atm × V = 1.5288 mol × 0.0821 L.atm/K.mol × 273.15 K

⇒V = 34.28 L ( 1.5*22.4 L)