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Roasting galena [lead(II) sulfide] is an early step in the industrial isolation of lead. How many liters of sulfur dioxide, measured at STP, are produced by the reaction of 7.07 kg of galena with 218 L of oxygen gas at 220°C and 2.00 atm? Lead(II) oxide also forms.

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nkirotedoreen46

Answer:

160.83 L, SOâ‚‚

Explanation:

From the question we are given;

Mass of PbS = 7.07 kg

Volume of Oxygen = 218 L

Temperature, T = 220°C or 493.15 K

Pressure, P = 2.00 atm

The equation for the reaction is;

2PbS + 3O₂ → 2PbO + 2SO₂

We are required to determine the volume of SOâ‚‚ formed.

We are going to use the following simple steps:

Step 1: Determine the number of moles of PbS

Number of moles = Mass ÷ Molar mass

Molar mass of PbS = 239.3 g/mol

Therefore;

Moles = 7070 g ÷ 239.3 g/mol

          = 29.54 moles, PbS

Step 2: Calculate the number of moles of Oxygen

To get the number of moles of oxygen gas we are going to use ideal gas equation.

PV = nRT, R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore, moles, n = PV ÷ RT

Moles of oxygen = (2.00 atm × 218 L) ÷ (0.082057 × 493.15K)

                            = 10.77 moles, O₂

  • Since, the number of moles of Oâ‚‚ are less compared to those of PbS, then oxygen is the rate limiting reagent.

Therefore, we can determine the number of moles of SOâ‚‚.

Step 3: Determining the number of moles of SOâ‚‚

From the equation, 3 moles of oxygen reacts with PbS to produce 2 moles of SOâ‚‚.

Therefore, with 10.77 moles of Oxygen;

Moles of SO₂= Moles of O₂× (2moles SO₂/3moles O₂)

= 10.77 moles × (2/3)

= 7.18 moles, SOâ‚‚

Step 4: Determine the volume of SOâ‚‚ produced

At STP 1 mole of a gas occupies a volume of 22.4 Liters

Therefore, 7.18 moles of SOâ‚‚ will have a volume of ;

= 7.18 moles × 22.4 L/mole

= 160.832 L

= 160.83 L

Thus, 160.83 L of SOâ‚‚ are produced by the reaction.

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