Question
Asked Dec 2, 2019
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Stomach acid is approximately 0.10 M HCl. How many mL of stomach acid can be neutralized by one regular antacid tablet that contains 500 mg of solid CaCO3 (100.9 g/mol) ?

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Expert Answer

Step 1

Given,

0.10 M HCl.

500 mg of soli...

Balanced reaction is given below,
2HC1 CaCO -> CaCl, + H,CO
According to the balanced equation, two moles of HCl reacts with one mole of calcium
carbonates gives one mole of calcium chloride and one mole carbonic acid
Molar mass of calcium carbonate is 100.9 g/mol
Number of moles of calcium carbonate is calculated as follows,
Mass
Moles
Molar mass
0.5g
Moles
100.9g/mol
Moles 4.955x 103
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Balanced reaction is given below, 2HC1 CaCO -> CaCl, + H,CO According to the balanced equation, two moles of HCl reacts with one mole of calcium carbonates gives one mole of calcium chloride and one mole carbonic acid Molar mass of calcium carbonate is 100.9 g/mol Number of moles of calcium carbonate is calculated as follows, Mass Moles Molar mass 0.5g Moles 100.9g/mol Moles 4.955x 103

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