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

34.28 L ( 1.5*22.4 L)

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:

Thus,

According to the reaction:-

4 moles of aluminum react with 3 moles of oxygen gas

1 mole of aluminum react with moles of oxygen gas

2.0384 moles of aluminum react with 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:

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)

the dependant variable is spilled water

the independent is the heat