The mass of 70%H2SO4 required for neutralization of 1mol of NaOH .
Question
The mass of 70%H2SO4 required for neutralization of 1mol of NaOH .
Solution
To solve this problem, we need to understand the reaction between Sodium Hydroxide (NaOH) and Sulfuric Acid (H2SO4). The balanced chemical equation is:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that 1 mole of H2SO4 reacts with 2 moles of NaOH. Therefore, 1 mole of NaOH will require 0.5 moles of H2SO4 for neutralization.
The molar mass of H2SO4 is approximately 98 g/mol. Therefore, the mass of H2SO4 required is 0.5 moles * 98 g/mol = 49 g.
However, the problem states that the H2SO4 is 70% by mass. This means that in 100 g of solution, there are 70 g of H2SO4. Therefore, to get 49 g of H2SO4, we need 49 g / 0.70 = 70 g of the 70% H2SO4 solution.
So, the mass of 70% H2SO4 required for neutralization of 1 mol of NaOH is 70 g.
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