Question

Enter your answer in the provided box. 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 5.05 kg of galena with 123 L of oxygen gas at 220°C and 2.00 atm

Answer 1

Answer : The volume of
`SO_2` produced are 90.7 liters.

Explanation :

The balanced chemical reaction will be:

`2PbS+3O_2\rightarrow 2PbO+2SO_2`

First we have to calculate the moles of PbS.

`\text{Moles of }PbS=\frac{\text{Mass of }PbS}{\text{Molar mass of }PbS}`

Molar mass of PbS = 239.26 g/mole

`\text{Moles of }PbS=(5.05kg)/(239.26)=(5.05* 1000g)/(239.26)=21.1mole`

Now we have to calculate the moles of
`O_2` by using ideal gas equation.

Using ideal gas equation :

`PV=nRT`

where,

P = Pressure of
`O_2` gas = 2.00 atm

V = Volume of
`O_2` gas = 123 L

n = number of moles
`O_2` = ?

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`O_2` gas =
`220^oC=273+220=493K`

Putting values in above equation, we get:

`2.00atm* 123L=n* (0.0821L.atm/mol.K)* 493K`

`n=6.07mole`

The number of moles of
`O_2` is, 6.07 mole

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 3 mole of
`O_2` react with 2 mole of
`PbS`

So, 6.07 moles of
`O_2` react with
`(6.07)/(3)* 2=4.05` moles of
`PbS`

From this we conclude that,
`PbS` is an excess reagent because the given moles are greater than the required moles and
`O_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`SO_2`

From the reaction, we conclude that

As, 3 mole of
`O_2` react to give 2 mole of
`SO_2`

So, 6.07 moles of
`O_2` react to give
`(6.07)/(3)* 2=4.05` moles of
`SO_2`

Now we have to calculate the volume of
`SO_2` produced at STP.

As, 1 mole of
`SO_2` contains 22.4 L volume of
`SO_2`

So, 4.05 mole of
`SO_2` contains
`4.05* 22.4=90.7L` volume of
`SO_2`

Therefore, the volume of
`SO_2` produced are 90.7 liters.