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.

160.83 L, SO₂

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 SO₂ formed.

We are going to use the following simple steps:

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

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

Therefore, we can determine the number of moles of SO₂.

From the equation, 3 moles of oxygen reacts with PbS to produce 2 moles of SO₂.

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, SO₂

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

Therefore, 7.18 moles of SO₂ will have a volume of ;

= 7.18 moles × 22.4 L/mole

= 160.832 L

= 160.83 L

Thus, 160.83 L of SO₂ are produced by the reaction.

the formula would be mgso4

mgso[subscript]4

the answer is true.