posted by aisha on .
what volume of 0.1 moldm-3 HCL acid would be required to dissolve 2.3 grams of calcium carbonate
equation CaCO3+2HCL----------=CaCl2+CO2+H20 (WATER)
Work out the Mr for CaCO3, which is 100.1gmol-1. Use this to work out the number of moles using the equation n=m/Mr, which is 2.3g/100.1gmol-1 = 0.02977 mol.
Next work out the ration of CaCO3 to HCl which is 1:2. Therefore the number of moles of HCL is twice that of CaCO3 [0.02977 mol X 2 = 0.05954].
Then use the equation Volume = Moles / Concentration, which is 0.05954 mol / 0.1 moldm-3 = 0.5954 dm3, your answer; the volume of HCl.
Hope that's right!! :D