Question

In: Chemistry

1. Assuming stomach acid is 0.1M HCl, how many grams of stomach acid would have been...

1. Assuming stomach acid is 0.1M HCl, how many grams of stomach acid would have been neutralized by your tablet? (use the average value)

2. Did the tablet neutralize at least 47 times its weight in stomach acid? ( use the average mass of your tablet)

show calculations

Average mass of tablet =2249.67mg and Average mass of Antacid = 807.33mg

Average mol of Antacid = 0.00807 and molar mass of antacid = 171.346

Solutions

Expert Solution

Ans. Balanced reaction: NaHCO3(s) + HCl(aq) ----------> NaCl(aq)+ H2O(l) + CO2(aq)

Stoichiometry: 1 mol NaHCO3 neutralizes 1 mol HCl.

#1. See stoichiometry, 1 mol NaHCO3 (antacid) neutralizes 1 mol HCl.

Thus,

Total moles of HCL neutralized = Moles of antacid consumed = 0.00807 mol

Given,

            Molarity of stomach acid = 0.1 M

Now, using,

            Number of moles of HCl = Molarity x Volume (V, in Liters) of stomach acid

            Or, 0.00807 mol = 0.1 M x V  

            Or, 0.00807 mol = (0.1 mol/ L) x V                                                ; [M = mol/ L]

            Or, V = 0.00807 mol / (0.1 mol/ L)

            Or, V = 0.0807 L

Hence, volume of stomach acid neutralized = 0.0807 L = 80.7 mL

Assuming the density of stomach acid to be equal to that of water (1.0 g/ mL), mass of stomach acid neutralized = Vol. of stomach acid x density

                                                = 80.7 mL x (1.0 g/ mL)

                                                = 80.7 g

#2. Mass of Antacid tablet = 2249.67 mg = 2.24967 g                          ; [1 g = 1000 mg]

Ratio of Mass of (Stomach acid / antacid tablet) = 80.7 g/ 2.24967 g

                                                = 35.87

That is, 1.0 gram antacid neutralizes 35.87 g of stomach acid.

Therefore,

The antacid does not neutralizes at least 47 times stomach acid of acid mass.


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