[acidburn191]

This question has been bugging mean because I can't even get a percentage between 50-100.
Please Help

Reaction equation for the dissolution of calcium carbonate in hydrochloric acid
CaCO₃ +2HCl --> CO₂ + H₂O+ CaCl₂

Reaction equation for the neutralization of the excess hydrochloric acid
HCl + NaOH --> NaCl + H₂O

Marble sample: 0.2548 g
Volume of HCl: 0.050L
Concentration of HCl: 0.1007 mol/L
Volume of NaOH: 0.02879
Concentration of NaOH: 0.02510 mol/L

Calculate the Percent CaCO₃ in the marble sample…

[Borek]

I am just guessing that it was a back titration.

Show what you did. I got something between 80 and 90%.

[Cherriyan]

As per my calculations, the percentage of calcium carbonate is coming out to be about 14. What value are you getting?

[DrCMS]

As per my calculations, the percentage of calcium carbonate is coming out to be about 14. What value are you getting?

You'd be wrong then because as Broek has already posted the correct answer is between 80 and 90%.

[Cherriyan]

Sorry! My previous answer was wrong as I interpreted the question incorrectly. But now I've got it.
From the reaction between HCl and NaOH, volume of excess HCl reacting with NaOH can be calculated (using normality equation, i.e., N₁V₁=N₂V₂). It's coming out to be about 0.00717 L.
Subtracting it from the total value of HCl given, we'll get the volume of HCl reacting with CaCO₃, which is 0.04282 L. This will lead us to the value of number of moles of HCl reacting with calcium carbonate, which is 0.00431. According to the equation,

CaCO₃ +2HCl --> CO₂ + H₂O+ CaCl₂

number of moles of CaCO₃ reacting = (no. of moles of HCl reacting with CaCO₃)/2 = 0.00215
=>weight of CaCO₃ = 0.21562g
=>percentage of CaCO₃ in marble = 84.62%

[Borek]

=>percentage of CaCO₃ in marble = 84.62%

You are aware of the fact that solving questions for others is against forum rules?