limestone (CaCO3) is used to remove acidic pollutants from smokestack flue gases. It is heated to form lime (CaO) , which reacts with sulfur dioxide to form calcium sulfite , assuming a 70% yield in the overall reaction , what mass of limestone required to remove all the sulfur dioxide formed by the combustion of 8.5x10^4 kg of coal that is 0.33 mass% sulfur?
mass coal is 8.5E4 kg or 8.5E7 g.
That is 0.33% S or 8.5E7 x 0.0033 = about 300,000 g but that's just an estimate. You should redo this and all calculations that follow to gt a more accurate answer. So how much SO2 is formed?
S + O2 ==> SO2
mols S = g/atomic mass = about 300,000/32 = about 10,000
Convert that to mols CaCO3.
CaCO3 ==> CaO
CaO + SO2 ==> CaSO3
10,000 mols S will use 10,000 mols
CaO which must have had 10,000 mols CaCO3
g CaCO3 = mols CaCO3 x molar mass CaCO3 = about 1,000,000 g or about 1,000 kg limestone. Post your work if you get stuck.
Well, well, well, it seems like we have a chemistry problem on our hands! Time to put on my lab coat and get silly with it!
First things first, let's find out how much sulfur dioxide is actually produced by that heap of coal. We know that coal is 0.33% sulfur, so we'll need to do a little calculation dance.
8.5x10^4 kg of coal multiplied by 0.0033 (0.33% in decimal form) gives us 280.5 kg of sulfur. Phew, that's a lot of stinky stuff! 🤢
Now, we need to figure out how much limestone we'll need to react with all that sulfur dioxide. The chemical equation tells us that one mole of sulfur dioxide reacts with one mole of limestone. But of course, we'll need some stoichiometry shenanigans to get the masses involved.
The molar mass of sulfur dioxide (SO2) is about 64 g/mol, and the molar mass of limestone (CaCO3) is around 100 g/mol. So, each mole of sulfur dioxide needs about 100 g of limestone to react with.
280.5 kg of sulfur dioxide is a whopping 280,500 g (remember, there are 1000 grams in a kilogram). So, dividing 280,500 g by the molar mass of sulfur dioxide (64 g/mol) gives us approximately 4392 moles of sulfur dioxide.
Now, since the overall reaction has a 70% yield, we only need to consider that portion. Multiplying 4392 moles by 0.70, we get around 3074 moles of sulfur dioxide.
So, if each mole of sulfur dioxide reacts with one mole of limestone, we need 3074 moles of limestone. But we're not done yet! We need to convert that into mass.
Multiplying 3074 moles by the molar mass of limestone (100 g/mol), we find that we need approximately 307,400 g (or 307.4 kg) of limestone.
There you have it! To remove all the sulfur dioxide produced by that coal, you'll need about 307.4 kg of limestone. Keep rockin' that chemistry! 🧪💥
To find the mass of limestone required to remove all the sulfur dioxide, we need to calculate the amount of sulfur dioxide formed by the combustion of coal, and then use stoichiometry to determine the amount of limestone needed to react with that amount of sulfur dioxide.
1. Determine the amount of sulfur in the coal:
Mass of coal = 8.5x10^4 kg
Mass% of sulfur in coal = 0.33%
Mass of sulfur in coal = (Mass% of sulfur / 100) x Mass of coal
= (0.33 / 100) x 8.5x10^4 kg
2. Determine the amount of sulfur dioxide produced from the amount of sulfur:
Molar mass of sulfur dioxide = 32.07 g/mol + 2 x 16.00 g/mol = 64.07 g/mol
Amount of sulfur dioxide produced = (Mass of sulfur / Molar mass of sulfur dioxide)
= (Mass of sulfur / 64.07 g/mol)
3. Determine the amount of limestone needed to react with sulfur dioxide:
From the balanced equation: CaCO3 + SO2 → CaSO3 + CO2
The stoichiometric ratio between limestone and sulfur dioxide is 1:1.
Amount of limestone needed = Amount of sulfur dioxide produced
4. Determine the mass of limestone needed:
Mass of limestone needed = Amount of limestone needed x Molar mass of limestone
Molar mass of limestone = 40.08 g/mol + 12.01 g/mol + 3 x 16.00 g/mol = 100.09 g/mol
Mass of limestone needed = Amount of limestone needed x 100.09 g/mol
Since the problem states a 70% yield in the overall reaction, multiply the mass of limestone needed by the reciprocal of the yield (1/0.70) to account for the yield:
Mass of limestone required = (Mass of limestone needed) / (0.70)
Now, we can substitute the values into the equations and calculate the mass of limestone required.
Mass of sulfur in coal = (0.33 / 100) x 8.5x10^4 kg
Amount of sulfur dioxide produced = (Mass of sulfur / 64.07 g/mol)
Amount of limestone needed = Amount of sulfur dioxide produced
Mass of limestone needed = Amount of limestone needed x 100.09 g/mol
Mass of limestone required = (Mass of limestone needed) / (0.70)
By performing the calculations, the mass of limestone required to remove all the sulfur dioxide is determined.
To determine the mass of limestone required to remove all the sulfur dioxide formed by the combustion of coal, we need to follow these steps:
Step 1: Calculate the mass of sulfur dioxide (SO2) formed by the combustion of coal.
First, we need to find the mass of sulfur (S) in the 8.5x10^4 kg of coal. Since the coal is 0.33% sulfur by mass, we can calculate:
Mass of sulfur = (0.33/100) * 8.5x10^4 kg = 280.5 kg
The balanced chemical equation for the combustion of sulfur can be written as:
S + O2 → SO2
From the equation, we can see that 1 mole of sulfur reacts with 1 mole of sulfur dioxide.
The molar mass of sulfur dioxide (SO2) is 32.06 g/mol.
Now, we can calculate the number of moles of sulfur dioxide (SO2) formed:
Moles of sulfur dioxide (SO2) = Mass of sulfur dioxide (g) / Molar mass of sulfur dioxide (g/mol)
Moles of sulfur dioxide (SO2) = (280.5 kg * 1000 g/kg) / 32.06 g/mol = 8750.93 mol
Step 2: Calculate the moles of limestone required.
Since the overall reaction has a 70% yield, we need to divide the moles of sulfur dioxide by the yield percentage to find the moles of limestone required:
Moles of limestone = Moles of sulfur dioxide / yield percentage
Moles of limestone = 8750.93 mol / (70/100) = 12501.33 mol
Step 3: Convert the moles of limestone to mass.
The molar mass of limestone (CaCO3) is 100.09 g/mol.
Now, we can calculate the mass of limestone required:
Mass of limestone = Moles of limestone * Molar mass of limestone
Mass of limestone = 12501.33 mol * 100.09 g/mol = 1,250,666.34 g
Therefore, approximately 1,250,666.34 grams (or 1,250.67 kg) of limestone is required to remove all the sulfur dioxide formed by the combustion of 8.5x10^4 kg of coal that is 0.33 mass% sulfur.