Answer:
16.2mL of 0.1111M NaOH are used
Explanation:
<em>volumetric flask of 250.0mL</em>
As first, the HCl reacts with CaCO3 as follows:
2HCl + CaCO3 → H2O + CaCl2 + CO2
<em>Where 2 moles of HCl react with 1 mole of CaCO3</em>
To solve this question we must find the moles of each reactant in order to determine the moles of HCl that remains:
<em>Moles HCl:</em>
0.01800L * (2.125mol / L) = 0.03825 moles HCl
<em>Moles CaCO3 -Molar mass: 100.09g/mol-</em>
0.4104g * (1mol / 100.09g) = 0.00410 moles CaCO3
Moles HCl that reacts:
0.00410 moles CaCO3 * (2 mol HCl / 1 mol CaCO3) =
0.00820 moles HCl
Moles HCl that remains:
0.03825 moles HCl - 0.00820 moles HCl:
<em>0.03005 moles HCl remains </em>
The HCl reacts with NaOH as follows:
HCl + NaOH → H2O + NaCl
That means, to reach the equivalence point, the moles of NaOH that must be added = Moles HCl, in a sample of 15.00mL, the moles of HCl are:
0.03005 moles HCl * (15.0mL / 250.0mL) = 0.001803 moles HCl = Moles NaOH. The volume of 0.1111M NaOH is:
0.001803 moles NaOH * (1L / 0.1111moles) = 0.016L 0.1111M NaOH are required, that is:
<h3>16.2mL of 0.1111M NaOH are used</h3>
