Question

**A piece of marble (assume it is pure CaCO3) reacts with 2.00L of 2.52 M HCl. After dissolution of the marble, a 10.00 mL sample of the resulting solution is withdrawn, added to some water, and titrated with 24.87 mL of 0.9987 M NaOH. What must have been the mass of the piece of marble?

Answer 1

Answer:

3.7 g

Explanation:

In this problem, marble reacted completely with an excess of HCl, which then reacted with NaOH

The reaction between marble and HCl is:

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

Then the excess HCl reacted with NaOH:

HCl + NaOH → NaCl + H₂O

So first let's calculate the amount of excess HCl that reacted with NaOH:

0.9987 M NaOH * 0.02487 L * $\frac{1molHCl}{1molNaOH}$ = 0.02483 mol HCl

Then we calculate the number of unreacted HCl moles in the original 2 L:

0.02483 mol HCl * 2 L / 0.01 L = 4.966 mol HCl

We substract that number from the initial HCl moles in order to find out how many of them reacted with the piece of marble

Initial HCl moles = 2.00 L * 2.52 M = 5.04 mol HCl

HCl moles that reacted with marble = 5.04 - 4.966 = 0.074 mol HCl

With that we can calculate the moles of marble, and finally the mass, using its molecular weight:

0.074 mol HCl * $\frac{1molCaCO_{3}}{2molHCl}$ *$\frac{100g}{1molCaCO_{3}}$ = 3.7 g marble