Pyrite, also known as fool's gold, is used commercially to produce SO₂ used in the production of paper products.
b. How many moles of SO₂ can be produced from 1.0 kg of pyrite?

Question

Pyrite, also known as fool's gold, is used commercially to produce SO₂ used in the production of paper products.

b. How many moles of SO₂ can be produced from 1.0 kg of pyrite?

Accepted Answer

Hello everyone today, we have the following problem. Carbon dioxide is one of the products during the combustion of a hydrocarbon. How many moles of carbon dioxide are produced from the combustion of 1.5 g kilograms of decade. So we know that combustion of a hydrocarbon is usually done in the presence of oxygen gas. So if we were to draw a reaction out, we would have our hydrocarbon here, our C 10 H 22 or decade and then it's going to react with our oxygen and that is going to form our carbon dioxide and our water. Now there is a balance this out. We need to see how many of each element we have on either side. So for carbons we have 10 on the left and one on the right. For hydrogen we have 22 on the left and two on the right, and for oxygen's we have two on the left and three on the right. So the first thing we want to balance is our carbons. So to get from 1 to 10, we simply need to add or multiply this 10, this carbon dioxide by 10. And notice that we also change the number of oxygen's. So instead of three, we have that 10 times that too, which is 20 plus that one we have 21 oxygen's on the right, so now we can balance our hydrogen To get from 2 to 11, we need to multiply this two x 11 so that we can get 22. Notice that we change the number of just again, So now that we have that 20 from that carbon dioxide and we have from the water. We now have 31 on our right hand side. And so what I like to do in this situation is In order to balance out the oxygen's on the left hand side. We're simply going to take the highest number of oxygen that we have, which is 31 and divide that by two. And so we can leave it as that fraction. And we can also note that for every one mole of decade, right? As a coefficient of one, we get 10 moles of carbon dioxide. So let's convert that kilograms of decades and decades. So we have our mass which is 1.5 kg of decade. To convert that. We need to first convert it to grams by using the conversion factor that one kg is equal to 10 to the third grams. We then can multiply by the molar mass of decade, which according to the periodic table for every one mole of decade, We have 142.3 g. And this comes from taking the individual masses of both hydrogen and the carbons. Multiplying it by how many there are and then adding those final numbers together. So when our units cancel out, we are left with 10.5 moles of our decking. Mhm. Then we want to determine the number of moles of C. O. To produce using these number of moles from decking. So We're gonna do is we're gonna take our given, we're gonna take our 10.5 moles a decade and we're gonna multiply by that multiple ratio. We said earlier that for every one mole of dead CAin that we use in the reaction, We produced 10 mils of carbon dioxide. Our units of decking cancel out. We are left with a final answer of 105 moles of carbon dioxide. And with that, we have answered our question overall, I hope this helped ahead until next time.

Pyrite, also known as fool's gold, is used commercially to produce SO₂ used in the production of paper products.
b. How many moles of SO₂ can be produced from 1.0 kg of pyrite?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Molar Mass

Chemical Reaction of Pyrite

Pyrite, also known as fool's gold, is used commercially to produce SO2 used in the production of paper products. b. How many moles of SO2 can be produced from 1.0 kg of pyrite?

Verified video answer for a similar problem:

Key Concepts

Stoichiometry

Molar Mass

Chemical Reaction of Pyrite

Pyrite, also known as fool's gold, is used commercially to produce SO₂ used in the production of paper products.
b. How many moles of SO₂ can be produced from 1.0 kg of pyrite?