Chemistry, 01.08.2020 08:01 alexis9263

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

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Answer from: Destinationz

Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.

The balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

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:

Al: 4 moles O₂: 3 moles Al₂O₃: 2 moles

Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted, the number of moles of Al that react is calculated as:

$14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles$

Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?

$amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}$

amount of moles of O₂= 0.405 moles

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?

$volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }$

volume= 9.072 L

Finally, 9.072 L of O₂ at STP would be required.

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Answer from: tystar84

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 require = 3 moles of O_2

Thus 0.54 moles of 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

Answer from: Quest

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Answer from: Quest

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