How many milligrams of NaHCO3 are in 500-mg tablet if 40.00 mL of 0.120 M HCl is required to neutralize the sample?
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The balanced chemical equation for the reaction of NaHCO3 and solution of HCl can be written as:
`NaHCO_3 + HCl -> NaCl + H_2O + CO_2`
To determine the amount of NaHCO3 present in the tablet, we shall employ stoichiometric procedures. Looking at the balanced chemical reaction, we can see that 1 mole of HCl is needed to react 1 mole of NaHCO3.
`Mol es HCl = (40.00)/(1000) L * 0.120 M = 0.004800 mol es HCl`
Finally, doing the stoichiometry:
`0.004800 mol es HCl * (1 mol e NaHCO_3)/(1 mol e HCl) * (84.007 grams)/(1 mol e NaHCO_3)`
`= 0.4032 grams NaHCO_3`
= 403.2 milligrams (mg) of NaHCO3 are present in the 500-mg tablet.
Note: 1000mg = 1 g
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