Question

4 Al + 3O2 → 2Al2O3 If 14.6 grams Al are reacted, how many liters of O2 at STP would be required?

Answer 1

Answer: 9.08 L

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles$

$4Al+3O_2\rightarrow 2Al_2O_3$

According to stoichiometry :

4 moles of $Al$ require  = 3 moles of $O_2$

Thus 0.54 moles of $Al$ will require=$\frac{3}{4}\times 0.54=0.405moles$  of $O_2$

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

$V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L$

Thus 9.08 L of $O_2$ at STP would be required