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9. When 11.0 g NaCl and 10.0 g H2SO4 are mixed and react according to theequation below, which is the limiting reagent?2NaCl + H2SO4 → Na2SO4 + 2HClA. NaClB. H2SO4C. Na2SO4D. HClE. No reagent is limitin

Question

  1. When 11.0 g NaCl and 10.0 g H2SO4 are mixed and react according to theequation below, which is the limiting reagent?2NaCl + H2SO4 → Na2SO4 + 2HClA. NaClB. H2SO4C. Na2SO4D. HClE. No reagent is limitin
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Solution

To determine the limiting reagent, we first need to calculate the number of moles of each reactant.

  1. Calculate the number of moles of NaCl: The molar mass of NaCl is approximately 58.44 g/mol. So, the number of moles of NaCl = mass / molar mass = 11.0 g / 58.44 g/mol = 0.188 mol

  2. Calculate the number of moles of H2SO4: The molar mass of H2SO4 is approximately 98.08 g/mol. So, the number of moles of H2SO4 = mass / molar mass = 10.0 g / 98.08 g/mol = 0.102 mol

  3. Compare the mole ratio: According to the balanced chemical equation, the mole ratio of NaCl to H2SO4 is 2:1. This means for every 2 moles of NaCl, we need 1 mole of H2SO4.

    However, we have less than 0.188/2 = 0.094 moles of H2SO4. Therefore, H2SO4 is the limiting reagent because it will be completely consumed before NaCl.

So, the answer is B. H2SO4.

This problem has been solved

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