Posted by **aisha** on Sunday, September 25, 2011 at 10:43am.

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)

- chemistry -
**Lyz-z**, Saturday, October 1, 2011 at 8:48am
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

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