Question

Fe₂O3 + 3CO=2Fe + 3CO₂ The reaction is carried out by mixing 130g of Fe₂O3 and 50g of CO i.Find the limiting reactant. ii.How much mass of iron is produced? iii.How many number of moles of unreacted reagent is left over? iv.Calculate the volume of CO₂ produced at 25°C and 2atm pressure.​

Answer 1

Step-by-step explanation:

i. To find the limiting reactant, we need to compare the number of moles of each reactant that we have. We can convert the mass of Fe2O3 and CO to moles by using their molar masses.

Fe2O3: 130g / (159.69 g/mol) = 0.817 mol

CO: 50g / (28.01 g/mol) = 1.78 mol

Since we have less moles of Fe2O3 than CO, Fe2O3 is the limiting reactant.

ii. To find the mass of iron produced, we first need to find the number of moles of Fe2O3 that react by using the balanced equation.

Fe2O3: 0.817 mol Fe2O3 x (2 mol Fe/1 mol Fe2O3) = 1.634 mol Fe.

then we can convert that to mass by using the molar mass of Fe.

1.634 mol Fe x (56 g/mol) = 91.984 g of Fe

iii. To find the number of moles of unreacted reagent, we need to subtract the number of moles of Fe2O3 that reacted from the original number of moles of Fe2O3 that we had.

0.817 mol - 0.817 mol = 0 mol of Fe2O3 left over.

iv. To find the volume of CO2 produced, we need to use the ideal gas law PV = nRT.

Where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is the temperature in kelvin.

Since we know the number of moles of CO2 produced is 1.634 mol and pressure and temperature in the question.

So, we can calculate the volume by using this information.

V = (1.634 mol * 8.314 J/mol*K * (25+273.15) K) / 2atm = 24.24 L

So, 24.24 L of CO2 is produced at 25°C and 2atm pressure.

Answer 2

i. The limiting reactant is Fe₂O₃. ii. The mass of iron produced is 77.99g. iii. There is no unreacted reagent left over. iv. The volume of CO₂ produced is 101.52 L.

Step-by-step explanation:

i. To find the limiting reactant, we need to compare the amounts of each reactant present and determine which one will be completely consumed first. Let's calculate the moles of Fe2O3 and CO:

Moles of Fe2O3 = mass / molar mass = 130g / (2 x 55.845g/mol + 3 x 15.999g/mol) = 0.6987 mol

Moles of CO = mass / molar mass = 50g / (12.01g/mol + 3 x 15.999g/mol) = 0.8266 mol

The ratio of Fe2O3 to CO in the balanced equation is 1:3. Since the molar ratio of Fe2O3 to CO is less than 1:3, Fe2O3 is the limiting reactant.

ii. To find the mass of iron produced, we can use the mole ratio between Fe2O3 and Fe:

Moles of Fe = moles of Fe2O3 x (2 mol Fe / 1 mol Fe2O3) = 0.6987 mol x 2 mol = 1.3974 mol

Mass of Fe = moles of Fe x molar mass = 1.3974 mol x 55.845g/mol = 77.99g

iii. Since Fe2O3 is the limiting reactant, all of the CO will be consumed. Therefore, there will be no unreacted reagent left over.

iv. To calculate the volume of CO2 produced, we need to use the ideal gas law equation:

Volume of CO2 = (moles of CO2 x 8.314 L/mol/K x temperature in Kelvin) / pressure

We need to find the moles of CO2 produced using the mole ratio between CO and CO2:

Moles of CO2 = moles of CO x (3 mol CO2 / 3 mol CO) = 0.8266 mol x 1 mol = 0.8266 mol

Now we can substitute the values into the ideal gas law equation, assuming a temperature of 25°C = 298K and pressure of 2 atm:

Volume of CO2 = (0.8266 mol x 8.314 L/mol/K x 298K) / 2 atm = 101.52 L