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
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
Solution
To determine the limiting reagent, we need to compare the number of moles of each reactant and their stoichiometric ratio in the balanced equation.
First, we need to calculate the number of moles of NaCl and H2SO4. To do this, we divide the given mass of each reactant by their respective molar masses.
The molar mass of NaCl is 58.44 g/mol, so the number of moles of NaCl is:
moles of NaCl = mass of NaCl / molar mass of NaCl = 11.0 g / 58.44 g/mol = 0.188 mol
The molar mass of H2SO4 is 98.09 g/mol, so the number of moles of H2SO4 is:
moles of H2SO4 = mass of H2SO4 / molar mass of H2SO4 = 10.0 g / 98.09 g/mol = 0.102 mol
Now, let's compare the moles of NaCl and H2SO4 to the stoichiometric ratio in the balanced equation.
From the balanced equation, we can see that the ratio of NaCl to H2SO4 is 2:1. This means that for every 2 moles of NaCl, we need 1 mole of H2SO4.
In our case, we have 0.188 moles of NaCl and 0.102 moles of H2SO4. Since the ratio of NaCl to H2SO4 is 2:1, we can see that we have an excess of NaCl.
Therefore, the limiting reagent is H2SO4.
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