Asked by Shariah
Calculate the mass f CaCo3, in Kg, needed to slurry to extract the SO2 present in 10m3 of industrial waste gases at rtp, if SO2 comprises 10% of this volume.
Answers
Answered by
DrBob222
10 m^3 = 10,000 L.
10% x 10,000 = 1000 L SO2
CaCO3 + SO2 ==> CaSO3 + CO2
Calculate n = mols SO2 in 1000 L.
PV = nRT
P = 1 atm
V = 1000 L
R = 0.08205
T = 298 K
Solve for n = number of mols SO2.
Convert that to mols CaCO3. mols CaCO3 - mols SO2 (look at the coefficients in the blaanced equation.).
Then grams CaCO3 = mols CaCO3 x molar mass CaCO3. Convert to kg. Post your work if you get stuck.
10% x 10,000 = 1000 L SO2
CaCO3 + SO2 ==> CaSO3 + CO2
Calculate n = mols SO2 in 1000 L.
PV = nRT
P = 1 atm
V = 1000 L
R = 0.08205
T = 298 K
Solve for n = number of mols SO2.
Convert that to mols CaCO3. mols CaCO3 - mols SO2 (look at the coefficients in the blaanced equation.).
Then grams CaCO3 = mols CaCO3 x molar mass CaCO3. Convert to kg. Post your work if you get stuck.
Answered by
riohni
okay thank you :)
Answered by
Briana
PV/RT = n
(1 x 1000)/(0.08205 x 298) = n
1000/24.45 = 40 moles in SO2 in 1000L
grams of CaCO3 = moles of CaCO3 X molar mass CaCO3
gram = 40g x 100g/mol
gram = 4000g
Convert to kg = 4kg
(1 x 1000)/(0.08205 x 298) = n
1000/24.45 = 40 moles in SO2 in 1000L
grams of CaCO3 = moles of CaCO3 X molar mass CaCO3
gram = 40g x 100g/mol
gram = 4000g
Convert to kg = 4kg
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