Calcium carbonate and sulfuric acid react according to the balanced reaction.
CaCO3+ H2SO4 = CaSO4 + CO2 + H2O
Starting from 300 grams of calcium carbonate with 3% impurities, what is the mass of calcium sulfate formed considering that the reaction is 90% yield?
300 g CaCO3 x 0.97 = approx 290 g real CaCO3 without the imputities but that's just a close number and you need calculate a more accurate number here and all the following steps.
mols CaCO3 = grams CaCO3/molar mass CaCO3.
Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaSO4. That is a 1:1 ratio in the equation, therefore, mols CaCO3 = mols CaSO4.
Now convert mols CaSO4 to grams. g = mols x molar mass and that gives you grams if the reaction is 100%. It isn't so grams under 100% x 0.90 = g with 90% yield. Post your work if you get stuck.
0.97% ??
To find the mass of calcium sulfate formed, we need to consider two factors: the amount of pure calcium carbonate and the reaction yield.
First, let's calculate the mass of pure calcium carbonate present in the 300 grams of the sample. Since the impurities account for 3%, the pure calcium carbonate will be 97% of the total mass (300 grams). We can calculate this as follows:
Pure calcium carbonate = (97/100) x 300 grams
Pure calcium carbonate = 291 grams
Now, we need to calculate the molar mass of calcium sulfate (CaSO4) and the molar mass of calcium carbonate (CaCO3) to compare the number of moles of each compound in the reaction.
Molar mass of CaSO4 = 40.08 g/mol (Ca) + 32.06 g/mol (S) + 4 * 16.00 g/mol (O)
Molar mass of CaSO4 = 136.14 g/mol
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 3 * 16.00 g/mol (C) + 3 * 16.00 g/mol (O)
Molar mass of CaCO3 = 100.08 g/mol
Next, we need to calculate the moles of pure calcium carbonate:
Moles of CaCO3 = Mass of pure CaCO3 / Molar mass of CaCO3
Moles of CaCO3 = 291 g / 100.08 g/mol
Moles of CaCO3 ≈ 2.909 mol
Using the balanced equation, we can determine the stoichiometry of the reaction, which tells us the mole ratio between calcium carbonate and calcium sulfate.
From the balanced equation, 1 mole of CaCO3 reacts to produce 1 mole of CaSO4.
Therefore, the moles of calcium sulfate formed = 2.909 mol.
Considering that the reaction yield is 90%, we multiply the moles of calcium sulfate by 0.90:
Effective moles of CaSO4 = Moles of CaSO4 x Reaction Yield
Effective moles of CaSO4 = 2.909 mol x 0.90
Effective moles of CaSO4 ≈ 2.618 mol
Finally, we calculate the mass of calcium sulfate formed:
Mass of CaSO4 = Effective moles of CaSO4 x Molar mass of CaSO4
Mass of CaSO4 = 2.618 mol x 136.14 g/mol
Mass of CaSO4 ≈ 356.8 grams
Therefore, the mass of calcium sulfate formed, considering a 90% reaction yield, would be approximately 356.8 grams.