A calculation that predicts how many milliliter of 0.121M of HCL acid solution needed to a 0.200g sample of calcium carbonate

Question

How many milliliters of 0.121 M of HCL solution you will need to titrate a 0.200g sample of calcuim carbonate

Answer 1

First, you must write a correctly balanced equation:

2HCl + CaCO3 ==> CaCl2 + H2O + CO2

Moles of CaCO3 present = 0.200 g x 1 mole/100.0 g = 0.002 moles

Moles HCl needed to react = 0.002 moles CaCO3 x 2 moles HCl/mole CaCO3 = 0.004 moles

mls of HCl needed:  (0.121 mol/L)(x L) = 0.004 moles 

x = 0.0331 mls