Chemistry, 16.03.2020 21:05 chancecharles9oug353

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 6.61 kg of galena with 154 L of oxygen gas at 220°C and 2.00 atm? Lead(II) oxide also forms.

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Answer from: Shami7457

Answer : The volume of sulfur dioxide measured at STP is, 113.6 L

Explanation :

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 = 239.27 g/mol

$\text{Moles of }PbS=\frac{6.61kg}{239.27g/mol}=\frac{6610g}{239.27g/mol}=27.62mol$

Now we have to calculate the moles of oxygen gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of O_2 gas = 2.00 atm

V = Volume of O_2 gas = 154 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:

$2.00atm\times 154L=n\times (0.0821L.atm/mol.K)\times 493K$

n=7.61mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$2PbS(s)+3O_2(g)\rightarrow 2PbO(s)+2SO_2(g)$

From the balanced chemical reaction we conclude that,

As, 3 moles of O_2 react with 2 moles of PbS

So, 7.61 moles of O_2 react with $\frac{2}{3}\times 7.61=5.07$ 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, 7.61 mole of O_2 react to give $\frac{2}{3}\times 7.61=5.07$ mole of SO_2

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

Using ideal gas equation:

PV=nRT

where,

P = Pressure of SO_2 gas = 1.00 atm

V = Volume of SO_2 gas = ?

n = number of moles SO_2 = 5.07 mol

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

T = Temperature of SO_2 gas = 273 K

Putting values in above equation, we get:

$1.00atm\times V=5.07mol\times (0.0821L.atm/mol.K)\times 273K$

V=113.6L

Thus, the volume of sulfur dioxide measured at STP is, 113.6 L

Answer from: Quest

i think it's c. i'm not sure but i think so.

Answer from: Quest

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.

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Roasting galena [lead(II) sulfide] is an early step in the industrial isolation of lead. How many li...