Question

1. A particular brand of antacid contains 500 mg of CaCO3 per 2.0 g tablet according to the label. How many mol of CaCO3 are contained in one tablet? nCaCO3 = ___ mol Ans. 0.005mol

2. The reaction by which the antacid neutralizes HCl is

2 HCl(aq) + CaCO₃(s) à CaCl₂(aq) + CO₂(g) + H₂O(l)

How many moles of HCl can be neutralized by one tablet?

n_HCl = 0.010 mol

3. 50.0 mL of 0.300 M HCl are used to dissolve a 2.00 g tablet. How many moles of acid are used to dissolve the tablet?

n_HCl = 0.015 mol

4.  After dissolving the 2.00 g tablet in 50.0 mL of 0.300 M HCl, the excess acid requires 53.13 mL of 0.100 M NaOH for the back titration. How many moles of excess acid were there in the 50 mL of HCl?

n_excess = ____ mol

5. (need help 4-7) How many mol of HCl were neutralized by the tablet?

n_neut = ___ mol

6. How many mol of CaCO₃ were there in the tablet?

n = ___ mol

7.How many mg of CaCO₃ were in the tablet?

mass = ___ mg

Show steps please

Answer 1

1. Mass of CaCO₃ = 500 mg = 0.5 g

Molar mass of CaCO₃ = 100 g/mol

Moles of CaCO₃ = Mass/ Molar mass = 0.5 g / (100 g/mol) =0.005 mol

2. From reaction

2 moles of HCl reacts with 1 mol of CaCO₃

so, Moles of CaCO₃ = 2*0.005 = 0.010 moles

3. Moles of HCl = Molarity * Volume = 0.3 M * 0.05L = 0.015 mol

4. Moles of NaOH = 0.100 M *0.05313 L = 0.005313 mol

Reaction between HCl and NaOH is:

HCl + NaOH ---> NaCl + H₂O

Moles of excess HCl = 0.005313 mol

5. Total moles of HCl added = 0.3 M * 0.050 L = 0.015 mol

Moles of HCl neutralized by tablet = 0.015 mol - 0.005313 mol = 0.009687 mol

6. Moles of CaCO₃ in tablet = Moles of HCl neutralized by tablet/ 2 = 0.009687/2 = 0.0048435 mol

7. Mass of CaCO₃ in tablet = 0.0048435 mol * 100 g/mol = 0.48435 g = 484.35 mg