How many moles of excess reactant when 338mL of 0.226M sulfuric acid reacts with 0.493L of 0.211M sodium hydroxide to form water and aqueous sodium sulfate?
To find the number of moles of the excess reactant, we need to determine the limiting reactant first. The limiting reactant is the one that will be completely consumed and determines the amount of product formed.
Let's calculate the number of moles for each reactant:
For sulfuric acid (H2SO4):
Volume of H2SO4 = 338 mL = 0.338 L
Concentration of H2SO4 = 0.226 M
Number of moles of H2SO4 = Volume x Concentration
= 0.338 L x 0.226 M
= 0.0765 moles
For sodium hydroxide (NaOH):
Volume of NaOH = 0.493 L
Concentration of NaOH = 0.211 M
Number of moles of NaOH = Volume x Concentration
= 0.493 L x 0.211 M
= 0.1037 moles
Now, we compare the mole ratio between H2SO4 and NaOH, which is 1:2.
Based on the mole ratio, we can see that for every 1 mole of H2SO4, we need 2 moles of NaOH. Therefore, the NaOH is the limiting reactant since there are only 0.1037 moles, while H2SO4 has 0.0765 moles.
To calculate the excess reactant, we subtract the number of moles of the limiting reactant from the number of moles of the other reactant.
Excess moles of H2SO4 = Moles of H2SO4 - Moles of NaOH
= 0.0765 moles - 0.1037 moles
= -0.0272 moles
Since we get a negative value, it means there is no excess H2SO4 left after the reaction. The NaOH is completely consumed, and the H2SO4 is the excess reactant.
Therefore, the number of moles of excess reactant (H2SO4) is 0 moles.