posted by Tony on .
In an experiment to find the percentage of calcium carbonate in sand from a beach, 1.86 g of sand reacted with an excess of dilute hydrochloric acid to give 0.55 g of carbon dioxide.
CaCO3 (s) + 2HCL (aq) – CaCl2(aq) + CO2(g) + H2O (l)
1.Calculate the number of moles of carbon present in 0.55 g of CO2
2.How many moles of calcium carbonate must have been present in the sand to produce this amount of carbon dioxide?
3. Calculate the mass of calcium carbonate present in the sand.
4. Calculate the percentage of calcium carbonate present in the sand.
Note to reader:
the questions i really find difficult is 2 and 3.
2) YOur balanced equation gives the answer: 1 mole of calcium carbonate yields one mole of carbon dioxide, which is 12 grams carbon. So for .55grams, you must have had .55/12 moles of calcium carbonate.
3. knowing the moles of calcium carbonate, change it to mass (grams).
What is the molarity of a solution containing 8 grams of solute in 500 mL of solution? Round to the nearest hundredth. Don't forget the units! (gram formula mass of solute = 24 g)