Question

Imagine if the HCl used in this experiment was prepared at a concentration of 0.150M. Using your calculated volume of HCl neutralized by your whole tablet, calculate the mass of active ingredient (CaCO3) in the tablet in milligrams. Show your work to receive credit.

Answer 1

Answer:

Explanation:

Given:

Concentration of HCl = 0.15 M

Volume of HCl neutralising the whole tablet, Va = 47.7 mL

Equation of the reaction:

CaCO3 + 2HCl --> CaCl2 + H2O + CO2

Number of moles = concentration × volume

= 0.15 × 47.7 × 10^-3

= 0.007155 mol

By stoichiometry, 1 mole of CaCO3 reacted completely with 2 moles of HCl. Therefore, number of moles of CaCO3 that occurred in the neutralisation reaction = 0.007155 mole of HCl × 1 moles of CaCO3/2 mole of HCl

= 0.0035775 moles of CaCO3

Molar mass of CaCO3 = 40 + 12 + (3 × 16)

= 100 g/mol.

Mass = molar mass × number of moles

= 100 × 0.0035775

= 0.35775 g

= 0.36 g of CaCO3.

Answer 2

Answer:

Molarity = moles / volume

moles of HCl       =  molarity ×volume

                            =  0.0150 M × 0.08189L

                             = 0.0123mol

∴ the amount of CaCO3 neutralized by 0.0123 mole of HCl will be in 1:2 ratio.

moles of CaCO3 = 0.0123 mol of HCl× 1mol CaCO3/ 2mol HCl

                             = 0 .00614 mol

∴ the tablet has 0.00614 mol of CaCO3.

Explanation:

Step 1:

In the above-experiment problem, HCl acid solution is basically treated with CaCO3 base active in the tablet.

The mass of CaCO3 present in the tablet must be obtained in milligram (mg).

First, write down the balanced reaction between HCl and CaCO3  

2Hcl aq + CaC03 s ⇒ CaCl2 aq + H2O l + C02 g

CaCO3 and HCl react in 1:2 ratio of moles respectively.

∴ for each mole of calcium carbonate there would be twice the moles of HCl  required for neutralization.

Step 2

the volume is  81.89 mL (i.e 0.08189 L) of 0.150 M HCl neutralizes the whole tablet containing unknown amount of CaCO3.

So, the number of moles of HCl present in the taken volume is calculated