Question

what mass of iron(iii) oxide is produced from excess iron metal and 6.8 l of oxygen gas at 102.5°c and 871 torr? 4 fe (s) 3 o2 (g) → 2 fe2o3 (s)

Answer 1

Approximately 97.18 grams of iron(III) oxide (Fe2O3) are produced from the reaction.

To calculate the mass of iron(III) oxide (Fe2O3) produced from the reaction of excess iron (Fe) and 6.8 L of oxygen gas (O2) at 102.5°C and 871 torr, use the ideal gas law to find the number of moles of oxygen gas and then use the stoichiometry of the balanced chemical equation to determine the amount of Fe2O3 produced.

Convert the given conditions to standard temperature and pressure (STP) to make the calculations easier. STP is 0°C (273.15 K) and 1 atm (760 torr). Use the combined gas law for this conversion:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:

- P1 = 871 torr

- V1 = 6.8 L

- T1 = 102.5°C + 273.15 (convert to Kelvin)

- P2 = 760 torr (STP)

- T2 = 0°C + 273.15 (STP in Kelvin)

First, convert the temperature to Kelvin:

T1 = 102.5°C + 273.15 = 375.65 K

T2 = 0°C + 273.15 = 273.15 K

Now, use the combined gas law to find the new volume at STP (V2):

(P1 * V1) / T1 = (P2 * V2) / T2

(871 torr * 6.8 L) / 375.65 K = (760 torr * V2) / 273.15 K

Now, solve for V2:

V2 = (871 torr * 6.8 L * 273.15 K) / (375.65 K * 760 torr)

V2 ≈ 32.99 L (rounded to two decimal places)

Now that you have the volume of oxygen gas at STP, you can calculate the moles of O2 using the ideal gas law:

PV = nRT

Where:

- P = pressure (in atm, use 1 atm at STP)

- V = volume (in liters)

- n = moles

- R = ideal gas constant (0.08206 L·atm/(mol·K))

- T = temperature (in Kelvin)

n(O2) = (1 atm * 32.99 L) / (0.08206 L·atm/(mol·K) * 273.15 K)

n(O2) ≈ 1.31 moles

Now, you can use the stoichiometry of the balanced chemical equation to find the moles of Fe2O3 produced. According to the balanced equation, 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.

Moles of Fe2O3 = (1.31 moles O2 * 2 moles Fe2O3) / 3 moles O2

Moles of Fe2O3 ≈ 0.87 moles Fe2O3

Now, calculate the molar mass of Fe2O3:

- Molar mass of Fe: 55.85 g/mol

- Molar mass of O: 16.00 g/mol

Molar mass of Fe2O3 = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 111.70 g/mol

Finally, calculate the mass of Fe2O3 produced:

Mass of Fe2O3 = Moles of Fe2O3 * Molar mass of Fe2O3

Mass of Fe2O3 ≈ 0.87 moles * 111.70 g/mol ≈ 97.18 grams

So, approximately 97.18 grams of iron(III) oxide (Fe2O3) are produced from the reaction.

Answer 2

Considering the ideal gas law and the reaction stoichiometry, 26.936 grams of iron(III) oxide is produced from excess iron metal and 6.8 L of oxygen gas at 102.5°c and 871 torr.

Reaction stoichiometry

The balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

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:

Fe: 4 moles

O₂: 3 moles

Fe₂O₃ : 2 moles

The molar mass of the compounds present in the reaction is:

Fe: 55.85 g/mole

O₂: 32 g/mole

Fe₂O₃ : 159.7 g/mole

Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:

Fe: 4 moles× 55.85 g/mole= 223.5 grams

O₂: 3 moles× 32 g/mole= 96 grams

Fe₂O₃ : 2 moles× 159.7 g/mole= 319.4 grams

Ideal gas law

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P×V = n×R×T

Mass of Fe₂O₃

You know for oxygen gas:

P= 871 torr

V = 6.8 L

n= ?

R= 62.364 torrL/molK

T= 102.5 °C= 375.5 K (being 0°C= 273 K)

Replacing on ideal gas law:

871 torr× 6.8 L = n×62.364 torrL/molK× 375.5 K

Solving;

(871 torr× 6.8 L) ÷ (62.364 torrL/molK× 375.5 K)= n

0.253 moles= n

Now, there is possible to aply the following rule of three: if by stoichiometry 3 moles of O₂ produce 319.4 grams of Fe₂O₃, if 0.253 moles of O₂ react how much mass of Fe₂O₃ will be formed?

mass of Fe₂O₃= (0.253 moles of O₂×319.4 grams of Fe₂O₃)÷3 moles of O₂

mass of Fe₂O₃= 26.935 grams

26.936 grams of iron(III) oxide is produced from excess iron metal and 6.8 L of oxygen gas at 102.5°c and 871 torr.