How many grams of sodium hydroxide will be produced when 23 gram of sodium reacts with wates at ordinary temperature ?
To determine the number of grams of sodium hydroxide produced when sodium reacts with water, we need to write and balance the chemical equation for the reaction:
2Na + 2H2O → 2NaOH + H2
The balanced equation tells us that for every 2 moles of sodium (Na), we will get 2 moles of sodium hydroxide (NaOH).
Now we need to calculate the number of moles of sodium from the given mass of 23 grams. We can do this using the molar mass of sodium (22.99 g/mol):
Moles of sodium (Na) = Mass of sodium (g) / Molar mass of sodium (g/mol)
Moles of sodium (Na) = 23 g / 22.99 g/mol = 1 mole
Since the balanced equation shows a 1:1 molar ratio between sodium and sodium hydroxide, we can conclude that 1 mole of sodium will produce 1 mole of sodium hydroxide.
Finally, we calculate the mass of sodium hydroxide using the molar mass of sodium hydroxide (40.00 g/mol):
Mass of sodium hydroxide (NaOH) = Moles of sodium hydroxide (mol) * Molar mass of sodium hydroxide (g/mol)
Mass of sodium hydroxide (NaOH) = 1 mole * 40.00 g/mol = 40 grams
Therefore, when 23 grams of sodium reacts with water, it will produce 40 grams of sodium hydroxide.