what is the theoretical yield of Sodium Hydroxide
You posted only part of a question.
To calculate the theoretical yield of Sodium Hydroxide (NaOH), you will need to know the balanced chemical equation and the amount of limiting reagent used.
The balanced chemical equation for the reaction involving Sodium Hydroxide is:
Na + H2O → NaOH + H2
Assuming you are given the amount of limiting reagent, let's say it is Na, you will need to convert the given amount of limiting reagent to moles.
Then, use the stoichiometry of the balanced equation to determine the mole ratio of NaOH to Na. In this case, it is 1:1, which means one mole of Na produces one mole of NaOH.
Finally, convert the mole of Na to grams of NaOH by using the molar mass of NaOH (the molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol).
It would be helpful if you could provide me with the specific amount or concentration of the limiting reagent in order to calculate the theoretical yield accurately.