Okay, so the question says that 1.5g of marble was dissolved in 100mL of 0.5M HCl. Then, there was excess acid that didn't react with the CaCO3 in the marble. We have to find out this amount to be able to tell how much HCl it took to dissolve the CaCO3, then to find out the amount of CaCO3 using the molar ratios.
Step 1: Titration of HCl and NaOH
HCl(aq) + (NaOH) -> H20(l) + NaCl(aq)
C1: 0.5M C2: 1M (concentrations, which are given)
V1: ? V2: 0.0245L (Volume - we are trying to find the volume of the HCl)
Since the molar ratios of HCl:NaOH are 1:1, we can use C1V1=C2V2
0.5 x ? = 1 x 0.0245
Volume of excess HCl = 0.049L
Since the excess HCl was 0.049L, the volume taken to dissolve the CaCO3 was 0.051L.
Now that we have the volume, we can find the number of moles of HCl used: 0.051 x 0.5(M) = 0.0255 moles
Since the CaCO3 was dissolved in the HCl, we have to look at the molar ratio:
CaCO3(s) + 2HCl(aq) -> H20(l) + CO2(g) + CaCl2(s)
CaCO3 is in a 2:1 ratio with HCl. That means that the number of moles of CaCO3 is half the number of moles of HCl used to dissolve it.
That value is 0.01275 moles.
Now we have to find the weight of this.
m = n.mm
= 0.01275 x (40.08+12.01+3x16.00)
= 1.2761g
As a percentage of the original 1.5g, this is 85.1%
