Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . Suppose 49. g of sulfuric acid is mixed with 18.8 g of sodium hydroxide. Calculate the minimum mass of sulfuric acid that could be left over by the chemical reaction. Round your answer to significant digits.
The reaction is: H2SO4 + 2NaOH → Na2SO4 + 2H2O
According to the stoichiometry, we have one mole of H2SO4 reacts with two moles of NaOH.
49g x 1mole / 98.0g = 0.5 moles H2SO4
18.8 x 1mole / 40.0g = 0.47 moles NaOH
Since two moles of NaOH is required to neutralize one mole of
H2SO4. But, in our case, we have less no. moles of NaOH.
Therefore, o.47 moles of NaOH neutralize 0.235 moles of H2SO4.
So, remaining no. moles of H2SO4 = 0.5 - 0.235 = 0.265 moles
Therefore, we have 0.265 moles of H2SO4 is unreacted in the system.
The mass of unreacted H2SO4 = mol. wt x no. moles = 98 g/mol x 0.265 mol = 25.97 g
hence, the mass of unreacted H2SO4 = 26 g
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