Question

A flask is filled with 50.0 ml of 0.15 M HCl acid. Exactly 0.054 grams of CaCo3 are placed in the acid,

neutralizing some of it. How many moles of NaOH must be added in a titration that neutralizes the rest

of the acid?

Answer 1

See explanation

Step-by-step explanation:

2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)

Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles

Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles

2 moles of HCl reacts with 1 mole of calcium carbonate

x moles of HCl reacts with 0.00054 moles of calcium carbonate

x = 2 * 0.00054/1

x = 0.00108 moles of HCl

Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles

Reaction of HCl and NaOH

HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)

Since the reaction is in the mole ratio of 1:1

0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH