Question

A Rolaids antacid tablet contains about .33 grams of NaAl(OH2)CO3. Calculate how much stomach acid, assumed to be 0.14 M HCl (aq) can be neutralized by one Rolaids tablet. The equation for neutralization reaction is

NaAl(OH)2CO3(s) + 4HCl (aq) -----> NaCl (aq) + AlCl3 (aq) + 3H2O (l) + CO2

Answer 1

Molar mass of NaAl(OH)₂CO₃ = 144 g/mol

So, 144 g of NaAl(OH)₂CO₃ = 1 mol

1 g of NaAl(OH)₂CO₃ = (1/144) mol

0.33 g of NaAl(OH)₂CO₃ = 0.33 x (1/144) mol = 0.0023 mol

So, one Rolaids tablet contains 0.0023 mol of NaAl(OH)₂CO₃.

Now,

NaAl(OH)₂CO₃(s) + 4HCl (aq) -----> NaCl (aq) + AlCl3 (aq) + 3H2O (l) + CO2

In this reaction

1 mole of NaAl(OH)₂CO₃ reacts with 4 moles of HCl.

0.0023 mole of NaAl(OH)₂CO₃ reacts with (4 x 0.0023) moles of HCl.

0.0023 mole of NaAl(OH)₂CO₃ reacts with 0.0092 moles of HCl.

0.14 M HCl means

0.14 moles of HCl in 1000 mL

1 moles of HCl in (1000 / 0.14) mL

1 moles of HCl in 7142.86 mL

0.0092 moles of HCl in (0.0092 x 7142.86) mL

0.0092 moles of HCl in 65.71 mL

So, 65.71 mL of 0.14 M HCl.