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.
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?
2 answers
0.97% ??
0 answers