What mass of hydrochloric acid (in grams) can be neutralized by 2.7 of sodium bicarbonate?
Here is a step by step procedure that will solve most stoichiometry problems.
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The equation you have is
NaHCO3 + HCl ==> H2O + CO2 + NaCl
To determine the mass of hydrochloric acid that can be neutralized by sodium bicarbonate, we need to consider the balanced chemical equation for the reaction between these two compounds.
The balanced chemical equation is as follows:
2 NaHCO3 + 2 HCl -> 2 NaCl + 2 H2O + 2 CO2
According to the equation, 2 moles of sodium bicarbonate (NaHCO3) react with 2 moles of hydrochloric acid (HCl).
To solve this problem, we can follow these steps:
Step 1: Find the molar mass of sodium bicarbonate (NaHCO3)
Molar mass of Na = 22.99 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of NaHCO3 = (22.99 g/mol) + (1.01 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 84.01 g/mol
Step 2: Calculate the number of moles of sodium bicarbonate (NaHCO3) using the given mass.
Moles = Mass / Molar mass
Moles of NaHCO3 = 2.7 g / 84.01 g/mol
= 0.0321 mol
Step 3: Calculate the number of moles of hydrochloric acid (HCl) using the stoichiometric ratio from the balanced equation.
According to the balanced equation, 2 moles of NaHCO3 react with 2 moles of HCl.
Therefore, the number of moles of HCl is the same as the number of moles of NaHCO3.
Moles of HCl = 0.0321 mol
Step 4: Calculate the mass of hydrochloric acid using the number of moles and molar mass.
Mass = Moles * Molar mass
Mass of HCl = 0.0321 mol * (1 mol/1) * (36.46 g/mol)
= 1.17 g
Therefore, 2.7 grams of sodium bicarbonate can neutralize approximately 1.17 grams of hydrochloric acid.