Posted by aisha on Sunday, September 25, 2011 at 10:43am.
what volume of 0.1 moldm3 HCL acid would be required to dissolve 2.3 grams of calcium carbonate
equation CaCO3+2HCL=CaCl2+CO2+H20 (WATER)

chemistry  Lyzz, Saturday, October 1, 2011 at 8:48am
Work out the Mr for CaCO3, which is 100.1gmol1. Use this to work out the number of moles using the equation n=m/Mr, which is 2.3g/100.1gmol1 = 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 moldm3 = 0.5954 dm3, your answer; the volume of HCl.
Hope that's right!! :D
Answer This Question
Related Questions
 chemistry  Calcium carbonate CaCO3 reacts with stomach acid (HCl, hydrochloric ...
 Chemistry  I would really aprreciate if someone could help me, thank you so ...
 Chemistry  When calcium carbonate is added to hydrochloric acid, calcium ...
 chemistry  Calcium carbonate reacts with hydrochloric acid according to the ...
 Chemistry 104  What mass (in grams) of calcium carbonate is needed to react ...
 Chemistry  Calcium carbonate reacts with HCl according to the following ...
 chemistry  CaCO3(s)+2HCl(aq)>CaCl2(aq)+H2O(l)+CO2 (g) How many grams of ...
 General Chemistry  When calcium carbonate is added to hydrochloric acid, ...
 Intro to Chemistry  When calcium carbonate is added to hydrochloric acid, ...
 Chemistry  When calcium carbonate is added to hydrochloric acid, calcium ...
More Related Questions