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Answer :
We need to determine the limiting reagent and compare the amount of sulfuric acid used to the initial mass of sulfuric acid. The minimum mass of sulfuric acid that could be left over by the chemical reaction is approximately 26.2 grams (rounded to two significant digits).
To calculate the minimum mass of sulfuric acid (H2SO4) that could be left over by the chemical reaction with sodium hydroxide (NaOH), we need to determine the limiting reagent and compare the amount of sulfuric acid used to the initial mass of sulfuric acid.
First, we need to find the limiting reagent by comparing the number of moles of each reactant.
Molar mass of H2SO4 = 2 * (1.01 g/mol (hydrogen) + 32.07 g/mol (sulfur) + 4 * 16.00 g/mol (oxygen)) = 98.09 g/mol
Molar mass of NaOH = 22.99 g/mol (sodium) + 16.00 g/mol (oxygen) + 1.01 g/mol (hydrogen) = 39.99 g/mol
Moles of H2SO4 = 72.69 g / 98.09 g/mol ≈ 0.741 moles
Moles of NaOH = 37.9 g / 39.99 g/mol ≈ 0.948 moles
From the balanced equation: H2SO4 + 2NaOH -> Na2SO4 + 2H2O, we can see that the stoichiometric ratio between H2SO4 and NaOH is 1:2.
Since the number of moles of H2SO4 (0.741 moles) is less than half the number of moles of NaOH (0.948 moles), H2SO4 is the limiting reagent.
To determine the minimum mass of H2SO4 left over, we use the stoichiometric ratio between H2SO4 and NaOH.
Moles of H2SO4 used = 0.741 moles
Moles of H2SO4 left over = Moles of H2SO4 initially - Moles of H2SO4 used
= 0.741 moles - 0.5 * Moles of NaOH
= 0.741 moles - 0.5 * 0.948 moles
≈ 0.741 moles - 0.474 moles
≈ 0.267 moles
Mass of H2SO4 left over = Moles of H2SO4 left over * Molar mass of H2SO4
= 0.267 moles * 98.09 g/mol
≈ 26.2 g
Therefore, the minimum mass of sulfuric acid that could be left over by the chemical reaction is approximately 26.2 grams (rounded to two significant digits).
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