the first step in the reaction of alka-seltzer with stomach acid consist of one

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 (aq) + HCl (aq) ---> NaCl (aq) + H2CO3(l)

1 mol                      1 mol

Moles ofHCl required for 2 mol NaHCO3

= ( 1 mol HCl/1 mol NaHCO3) * 2 mol NaHCo3 = 2 mol HCl

So we need only 2 mol HCl to react with 2 mol NaHCO3 but we have 8 mol HCl which are in excess

So NaHCO3 is limiting reactant

Mols of H2CO3 = (1 mol H2CO3 / 1 mol NaHCO3) * 2 mol NaHCO3

= 2 mol H2CO3 (answer)

Excess reactantr (HCl) left = molesgiven

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