. What mass of NaOH (s) is required to react exactly with 25 ml of 0.3M H2SO4 (aq)?
To determine the mass of NaOH required to react with 25 ml of 0.3M H2SO4, we need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between NaOH and H2SO4 is:
2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
In this equation, we can see that 2 moles of NaOH react with 1 mole of H2SO4.
First, we need to calculate the number of moles of H2SO4 present in 25 ml of 0.3M H2SO4 solution.
Moles of H2SO4 = (concentration * volume) / 1000
= (0.3 * 25) / 1000
= 0.0075 moles
According to the balanced chemical equation, we require 2 moles of NaOH to react with 1 mole of H2SO4. Therefore, the moles of NaOH required would be twice the moles of H2SO4.
Moles of NaOH required = 2 * 0.0075
= 0.015 moles
Finally, to calculate the mass of NaOH, we can use the molar mass of NaOH, which is 22.99 g/mol.
Mass of NaOH = Moles of NaOH * Molar mass of NaOH
= 0.015 * 22.99
= 0.34485 grams
Therefore, approximately 0.34485 grams of NaOH is required to react exactly with 25 ml of 0.3M H2SO4.