Final answer:
The mass of sodium in the mixture is 32 g.
Step-by-step explanation:
To find the mass of sodium in the mixture, we need to understand the reaction that occurs when sodium reacts with water. The balanced equation for this reaction is: 2Na + 2H2O --> 2NaOH + H2.
From the equation, we can see that 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas. This means that the mole ratio between sodium and sodium hydroxide is 1:1.
In the given problem, the solution obtained from the reaction with water is neutralized with 517.3 ml of 1.0 M H2SO4. In order for a full neutralization to occur, twice as many moles of sodium hydroxide must react with the H2SO4. Therefore, we have 1 mole of sodium hydroxide reacting with 1 mole of sodium in the mixture.
Given the molar mass of sodium (Na) as 23 g/mol, we can set up the following equation to find the mass of sodium:
(32 g Na) / (23 g/mol Na) = (x moles Na) / (1 mole Na)
Solving for x gives us x = 1.391 moles, which is equal to the number of moles of sodium hydroxide.
Therefore, the mass of sodium in the mixture is 1.391 moles * 23 g/mol = 32.013 g. Rounding to the nearest gram, the mass of sodium is 32 g.