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

Answer:

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{55\ g}{26.981539\ g/mol}$

$Moles= 2.0384\ mol$

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 $\frac{3}{4}$ moles of oxygen gas

2.0384 moles of aluminum react with $\frac{3}{4}\times 2.0384$ 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)