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Caltrain tablets contains calcium carbonate. If one tablet requires 34.46 mL of 0.291 M solution of hydrochloric acid for neutralization, how many milligrams of calcium carbonate is in this caltrate tablet?

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There are 500 mg of calcium carbonate is in this caltrate tablet

The equation of the reaction is;

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

Next, we have to obtain the amount HCl reacted as follows;

Number of moles = concentration × volume

Volume = 34.46 mL or 0.03446 L

Concentration =  0.291 M

Number of moles = 0.291 M ×  0.03446 L

Number of moles = 0.01 moles

Since the reaction of HCl and CaCO3 are in the ratio of 2:1

2 moles of HCl reacts with 1 mole of CaCO3

0.01 moles of HCl reacts with  0.01 moles ×  1 mole/2 moles

= 0.005 moles of CaCO3

Mass of CaCO3 = number of moles  × molar mass

Mass of CaCO3 =  0.005 moles of CaCO3 × 100 g/mol

Mass of CaCO3 =  0.5 g or 500 mg

Learn more: https://brainly.com/question/9743981

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