The way I approached this problem was:
1. find number moles NaOH from the 36 grams of Na, and I got about 1.57 mol NaOH
2. find number of moles of NaOH from 53 mL of H2O, and I got 0.00237 mol NaOH
Therefore, H2O is limiting reactant since it created less amount of NaOH using all of it.

To calculate how much Na (excess reactant) is used in experiment, take 0.00237 moles NaOH and use stoichiometry to get grams of Na (0.0544g Na).

To calculate how much of Na was left after reaction do 36-0.0544=35.9 g Na remaining after reaction.

Is there a better way to do this? Like if I saw how much Na is used in reaction with 53 mL of H20 and vice versa?

According the chemical reaction you can see that 2 Na consumed also 2 H2O, so the ratio is 1:1.
Calculate the moles of each you find you have 1,565 mol Na and 2,944 mol H2O.
This has no ratio of 1: 1. It means all  sodium is consumed some water is left. The difference gives 1,379 mol what reflects to 24,83 g or ml.

Quote from: ann on November 29, 2022, 10:30:23 PM
The way I approached this problem was:


To calculate how much Na (excess reactant) is used in experiment, take 0.00237 moles NaOH and use stoichiometry to get grams of Na (0.0544g Na).

To calculate how much of Na was left after reaction do 36-0.0544=35.9 g Na remaining after reaction.

Is there a better way to do this? Like if I saw how much Na is used in reaction with 53 mL of H20 and vice versa?

you did error in calculations.
convert g of Na ->. moles of Na ----> moles of NaOH
Convert mL of water -----> mass of water (using density) ------> moles H2O -----> moles Of NaOH
Check which one is giving you minimum moles of NaOH is the limiting reactant.
Mass of LR = 0 (finally)
Mass of NaOH formed is from LR
How much excess left?
grams LR ---> moles LR -----> moles of excess -----> grams of excess.
This is the amount reacted and left is given - reacted mass.