Question

A 0.205 g sample of CaCO3 (Mr = 100.1 g/mol) is added to a flask along with 7.50 mL of 2.00 M HCl. CaCO3(aq) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)Enough water is then added to make a 125.0 mL solution.

Answer 1

The volume of the NaOH used is calculated as 14 mL.

What is stoichiometry?

The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.

Here;

Number of  moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles

Number of  moles of HCl = 2.00 M * 7/1000 L = 0.014 moles

2 moles of HCl reacts with 1 mole of CaCO3

x moles of HCl reacts with 0.00205 moles of CaCO3

x = 0.00205 moles * 2/1 = 0.0041 moles

Hence HCl is the excess reactant

Amount of excess HCl =  0.014 moles -  0.0041 moles = 0.0099 moles

Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M

Using;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

VB = CAVANB/CBNA

VB = 0.0792 M * 10 mL * 1/ 0.058 M

VB = 14 mL

Missing parts;

A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?

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