Final answer:
0.0452 grams of hydrochloric acid can be neutralized by 2.6 grams of sodium bicarbonate.
Step-by-step explanation:
To determine the mass of hydrochloric acid that can be neutralized by 2.6 g of sodium bicarbonate, we need to write a balanced chemical equation for the reaction between aqueous sodium bicarbonate and aqueous hydrochloric acid:
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(1) + CO2(g)
From the equation, we can see that the mole ratio of NaHCO3 to HCl is 1:1. Therefore, 2.6 g of NaHCO3 would react with an equal molar amount of HCl.
To convert the mass of NaHCO3 to moles, we divide by its molar mass:
2.6 g NaHCO3 / (22.99 g/mol + 1.01 g/mol + 12.01 g/mol + 3(16.00 g/mol)) = 0.0452 mol NaHCO3
Since the mole ratio of NaHCO3 to HCl is 1:1, the mass of HCl that can be neutralized is also 0.0452 g.