Question

If you mix 25.0 mL of 0.10 M HCL with 0.055g of calcium carbonate, how many moles of HCL are not neutralized by the calcium carbonate

Answer 1

The chemical equation representing the reaction of calcium carbonate with hydrochoric acid is:

`CaCO_(3)(s)+2HCl(aq)-->CaCl_(2)(aq)+H_(2)O(l)+CO_(2)( g)`

Moles of HCl given =
`25.0mL*(1L)/(1000mL)* (0.10mol)/(L) =0.0025molHCl`

Moles of
`CaCO_(3)`=
`0.055 g *(1mol)/(100.09g) =5.50*10^(-4)molCaCO_(3)`

Moles of HCl that would react with
`5.50*10^(-4)molCaCO_(3)`:

`5.50*10^(-4)molCaCO_(3)* (2mol HCl)/(1 mol CaCO_(3)  )`=0.0011mol HCl

Moles of HCl that would not be neutralized = 0.0025 mol - 0.0011 mol

= 0.0014 mol HCl