The first step in the reaction of Alka-Seltzer with stomach acid consists of one

Question

The first step in the reaction of Alka-Seltzer with stomach acid consists of one mole of sodium bicarbonate (NaHCOa) reacting with one mole of hydrochloric acid (HCI) to produce one mole of carbonic acid (H_2CO_3), and one mole of sodium chloride (NaCI). Using this chemical stoichiometry, determine the number of moles of carbonic add that can be produced from 3 mol of NaHCO_3 and 9 mol of HCI. Which of the two reactants limits the number of moles of H_2CO_3 that can be made? How much excess reactant remains after reaction?

Explanation / Answer

NaHCO3 + HCL ----> NaCl + H2CO3

1 mole of HCL can produce 1 mole of carbonic acid

since we have only 3 moles of NaHCO3 which is consumed first and acts as a limiting reagent so we get only 3 moles of H2CO3 along with 6 mole of HCl remaining as excess reactant

as excess reactant remaining in the system

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