Ca(OH)2 + SO2 ----> CaSO3 + H2O
Assuming that this process is only 80 % efficient, how much Ca(OH)2 would be required to remove 1000 grams of SO2?
Ca(OH)2 + SO2 ----> CaSO3 + H2O
mols SO2 = 1000/molar mass SO2 = approx 15.6 but you need to confirm that as well as all of the other numbers I do becuase I estimate some of them.
Convert to mols Ca(OH)2 and
15.6 mols SO2 x (1 mol Ca(OH)2/1 mol SO2) = 15.7.
Grams Ca(OH)2 = mols x molar mass = approx 15.7 x about 74 = about 1160 g if we are talking 100% efficient.
If only 80%, then it will take 1160/0.80 = ?
To calculate the amount of Ca(OH)2 required to remove 1000 grams of SO2, we need to consider the stoichiometry of the reaction and the given efficiency.
The balanced chemical equation for the reaction is:
Ca(OH)2 + SO2 → CaSO3 + H2O
According to the equation, we can see that the stoichiometric ratio between Ca(OH)2 and SO2 is 1:1. This means that for every 1 mole of SO2, we need 1 mole of Ca(OH)2.
To calculate the moles of SO2, we need to convert the given mass of 1000 grams to moles using the molar mass of SO2:
Molar mass of SO2 (32.07 g/mol) = 2(16.00) + 32.07 = 64.07 g/mol
Moles of SO2 = grams of SO2 / molar mass of SO2
Moles of SO2 = 1000 g / 64.07 g/mol
Moles of SO2 ≈ 15.61 mol
Since the efficiency is stated to be 80%, we need to calculate the actual amount of Ca(OH)2 required by dividing the moles of SO2 by the efficiency:
Actual moles of Ca(OH)2 = moles of SO2 / efficiency
Actual moles of Ca(OH)2 = 15.61 mol / 0.80
Actual moles of Ca(OH)2 ≈ 19.51 mol
Finally, to calculate the required mass of Ca(OH)2, we multiply the actual moles of Ca(OH)2 by its molar mass:
Mass of Ca(OH)2 = actual moles of Ca(OH)2 * molar mass of Ca(OH)2
Mass of Ca(OH)2 ≈ 19.51 mol * (40.08 + 16.00 + 1.01 + 1.01) g/mol
Mass of Ca(OH)2 ≈ 19.51 mol * 58.09 g/mol
Mass of Ca(OH)2 ≈ 1131.66 g
Therefore, approximately 1131.66 grams of Ca(OH)2 would be required to remove 1000 grams of SO2, assuming an 80% efficiency.
To determine the amount of Ca(OH)2 required to remove 1000 grams of SO2, we need to consider the stoichiometry of the balanced chemical equation and take into account the 80% efficiency.
The balanced chemical equation is:
Ca(OH)2 + SO2 → CaSO3 + H2O
According to the stoichiometry of the equation, for every 2 moles of Ca(OH)2, 1 mole of SO2 is reacted. This means that the molar ratio between Ca(OH)2 and SO2 is 2:1.
To find the number of moles of SO2 in 1000 grams of SO2, we need to use the molar mass of SO2, which is approximately 64.06 g/mol. Therefore, the number of moles of SO2 in 1000 grams is:
moles of SO2 = mass of SO2 / molar mass of SO2
moles of SO2 = 1000 g / 64.06 g/mol ≈ 15.60 moles of SO2
Since the molar ratio between Ca(OH)2 and SO2 is 2:1, the number of moles of Ca(OH)2 required would be half of the number of moles of SO2:
moles of Ca(OH)2 required = moles of SO2 / 2
moles of Ca(OH)2 required = 15.60 moles of SO2 / 2 ≈ 7.80 moles of Ca(OH)2
Now, since the process is only 80% efficient, we need to calculate the amount of Ca(OH)2 required, taking into account the efficiency.
Actual moles of Ca(OH)2 required = moles of Ca(OH)2 / efficiency
Actual moles of Ca(OH)2 required = 7.80 moles of Ca(OH)2 / 0.80 ≈ 9.75 moles of Ca(OH)2
Finally, to convert moles of Ca(OH)2 into grams, we multiply by the molar mass of Ca(OH)2, which is approximately 74.10 g/mol:
grams of Ca(OH)2 required = moles of Ca(OH)2 required * molar mass of Ca(OH)2
grams of Ca(OH)2 required = 9.75 moles of Ca(OH)2 * 74.10 g/mol ≈ 722.32 grams of Ca(OH)2
Therefore, approximately 722.32 grams of Ca(OH)2 would be required to remove 1000 grams of SO2, assuming a 80% efficiency.