the first step in the reaction of alka-seltzer with stomach acid consist of one
ID: 809255 • Letter: T
Question
the first step in the reaction of alka-seltzer with stomach acid consist of one mole of sodium bicarbonate (NaHCO3) reacting with one mole of hydrochloric acid(HCl) to produce one mole of carbonic acid(H2CO3) and one mole of sodium chloride(NaCl) ... using this chemical stiochimetry determine the number of moles of carbonic acid that can be produced from 2 mole of NaHCO3 and 8 mole of HCl............ mole of H2CO3 ? which of the two reactant limits the number of moles of H2CO3 than can be made, HCL or NaHCO3? how much excess reactant remin after the reaction?
Explanation / Answer
NaHCO3 + HCl ? H2CO3 + NaCl
4 moles of NaHCO3 would react completely with 4 moles of HCl, but there is more HCl present than that, so HCl is in excess and NaHCO3 is the limiting reactant.
4 moles of NaHCO3 produce 4 moles of H2CO3.
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