A particular brand of antacid contains 500 mg of CaCO_3 per 2.0 g tablet accordi

Question

A particular brand of antacid contains 500 mg of CaCO_3 per 2.0 g tablet according to the label. How many moles of CaCO_3 are in one tablet? The reaction by which the antacid neutralizes HCl is 2 HCl (aq) + CaCO_3 (s) rightarrow CaCl_2 (aq) + CO_2 (g) + H_2 O (l) How many moles of HCl can be neutralized by one tablet? 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? The excess acid then 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? How many moles of HCl were neutralized by the tablet? How many moles of CaCO_3 were in the tablet? How many mg?

Explanation / Answer

(1) Mass of CaCO3 per tablet = 500 mg = 500 / 1000 g = 0.500 g.

Number of moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3 = 0.500 / 100 = 0.00500 mol

n = 5.00 * 10-3 mol

(2)

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

From the balanced equation,

1 mol of CaCO3 can neeutralise 2 mol of HCl

then, 0.00500 mol of CaCO3 can neutralise 2 * 0.00500 = 0.0100 mol of HCl

(3)

Moles of acid used = Molarity * Volume os solution in mL / 1000 = 0.300 * 50.0 / 1000 = 0.0150 mol

(4)

Moles of NaOH used = 0.100 * 53.13 / 1000 = 0.00531 mol

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

From the above equation,

Moles of NaOH = moles of HCl = 0.00531 mol

(5)

Moles of HCl neutralised by the tablet=Moles of HCl - excess of moles of HCl remained=0.00969mol

(6)

Moles of CaCO3 = (1/2) * moles of HCl neutralised = 0.00969 / 2 = 0.00484 mol

Mass of CaCO3 = 0.00484 * 100 = 0.484 g. = 484 mg.

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