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charle [14.2K]
3 years ago
6

In a blast furnace, iron(III) oxide is used to produce iron by the following (unbalanced) reaction: Fe2O3 + CO---------->Fe +

CO2 a) If 4.00 kg FeO3 are available to react, how many moles of CO are needed? b) How many moles of each product are formed?
Chemistry
1 answer:
rewona [7]3 years ago
4 0

Answer:

a) The no. of moles of CO are needed = 75.15 mol.

b) The no. of moles of Fe are produced = 50.1 mol.

The no. of moles of CO₂ are needed = 75.15 mol.

Explanation:

  • Firstly, we need to write the balanced reaction:

<em>Fe₂O₃ + 3CO → 2Fe + 3CO₂,</em>

It is clear that 1 mol of Fe₂O₃ reacts with 3 mol of CO to produce 2 mol of Fe and 3 mol of CO₂.

<em>a) If 4.00 kg Fe₂O₃ are available to react, how many moles of CO are needed?</em>

  • We need to calculate the no. of moles of 4.00 kg Fe₂O₃:

n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

<u><em>Using cross multiplication:</em></u>

1 mol of Fe₂O₃ need  → 3 mol of CO to react completely, from stichiometry.

25.05 mol of Fe₂O₃ need  → ??? mol of CO to react completely.

∴ The no. of moles of CO are needed = (3 mol)(25.05 mol)/(1 mol) = 75.15 mol.

<em>b) How many moles of each product are formed?</em>

  • To find the no. of moles of Fe formed:

<em><u>Using cross multiplication:</u></em>

1 mol of Fe₂O₃ produce  → 2 mol of Fe, from stichiometry.

25.05 mol of Fe₂O₃ produce  → ??? mol of Fe.

∴ The no. of moles of Fe are produced = (2 mol)(25.05 mol)/(1 mol) = 50.1 mol.

  • To find the no. of moles of CO₂ formed:

<u><em>Using cross multiplication:</em></u>

1 mol of Fe₂O₃ produce  → 3 mol of CO₂, from stichiometry.

25.05 mol of Fe₂O₃ produce  → ??? mol of CO₂.

∴ The no. of moles of CO₂ are needed = (3 mol)(25.05 mol)/(1 mol) = 75.15 mol.

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Answer:

P_{He}=9.06torr

Explanation:

Hello there!

In this case, we can identify the solution to this problem via the Dalton's rule because the partial pressure of helium is given by:

P_{He}=x_{He}P_T

Whereas the mole fraction of helium is calculated by firstly obtaining the moles and then the mole fraction:

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