An antacid preparation of Na2CO3 claims that a 3.00 g tablet contains enough antacid to neutralize full stomach acid (0.100 M HCl). Assuming that the average stomach contains 0.750 L of acid, calculate the moles of Na2CO3 in the antacid. Refute or defend the claim.
1 Answer | Add Yours
The reaction equation of the acid in the stomach and the antacid can be written as:
`Na_2CO_3 + 2 HCl -> 2 NaCl + H_2O + CO_2`
To know if the prepared antacid is able to neutralize the acid in the stomach, the amount of Na2CO3 should be greater than or equal to the amount of moles of acid produced in the stomach.
`Mol es of acid (HCl) = (0.750 L )*( 0.1 M) = 0.075 mol es HCl`
`0.075 mol es HCl * (1 mol e Na_2CO_3)/(2 mol es HCl) * (105.99 grams Na_2CO_3)/(1 mol e Na_2CO_3)`
= 3.975 grams Na2CO3
Answer: the average stomach needs 3.975 grams of antacid in order to completely neutralize the stomach. The prepared antacid is only 3.00 grams and is obviously not enough to neutralize the acid completely.
We’ve answered 333,521 questions. We can answer yours, too.Ask a question