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

Atomicity occurs to prevent updates of the database occurring only partially, which can cause greater problems than rejecting the whole series outright.some are monatomic, diatomic, triatomic etc because it shows the number of atoms. for example, monatomic means it contains only one atom.

Answer from: Quest

answer carbon and three oxygen atoms;

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