Question

Расчитайте объём затраченного кислорода на реакцию со 112г. железа, если результат этой реакции железная окалина (Fe3O4)

Answer 1

3Fe+2O2 = Fe3O4.
1) n(Fe) = m/Mr =11,2/55,8=0,2 моль.
2) 3 моль - 2 моль (по ур-ю)
    0,2 моль - x моль (по усл)
   x=0,133.
3) n(O2)=V/Vm => V=Vm*n=22,4*0,133=2,98 л.

Ответ: 2,98 л.

Answer 2

Calculation of Oxygen Volume for the Reaction with Iron

To calculate the volume of oxygen consumed in the reaction with 112g of iron, we need to determine the stoichiometry of the reaction. The balanced equation for the reaction between iron and oxygen to form iron(II,III) oxide (Fe3O4) is:

3Fe + 2O2 → Fe3O4

From the balanced equation, we can see that 3 moles of iron react with 2 moles of oxygen to produce 1 mole of iron(II,III) oxide.

To calculate the moles of iron, we need to know the molar mass of iron. The molar mass of iron (Fe) is approximately 55.845 g/mol.

Let's calculate the moles of iron:

Moles of iron = mass of iron / molar mass of iron

Moles of iron = 112g / 55.845 g/mol

Moles of iron ≈ 2.003 mol

According to the stoichiometry of the reaction, 3 moles of iron react with 2 moles of oxygen. Therefore, the moles of oxygen required can be calculated using the ratio:

Moles of oxygen = (moles of iron / 3) * 2

Moles of oxygen = (2.003 mol / 3) * 2

Moles of oxygen ≈ 1.335 mol

Now, to calculate the volume of oxygen, we need to use the ideal gas law equation:

PV = nRT

Where: P = pressure (assume standard pressure, 1 atm) V = volume of gas n = moles of gas R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (assume room temperature, around 298 K)

Rearranging the equation, we get:

V = (n * R * T) / P

Substituting the values:

V = (1.335 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1 atm

V ≈ 33.06 L

Therefore, the volume of oxygen consumed in the reaction with 112g of iron is approximately 33.06 liters.

Please note that the above calculation assumes ideal conditions and may not account for any deviations from ideal behavior.