How do I find delta S of 2 Na(s) + 2H2O(l) -> 2NaOH(aq) +H2(g)?

I tried to find it using this....
^Srxn= 2S(NaOH) - S(H2) -2S(Na) - 2S(H2O)
= 2(48.1)-130.7 - 2(51.21) - 2(69.95)
= 96.2 - 130.7 - 102.42 - 139.9
= -279.82
Where did I go wrong??

My tables aren't quite the same as your (but I have an old old text). But

dSrxn = (n*dSproducts) - (n*dSreactants)
dSrxn = (2*NaOH + H2) - (2*Na + 2H2O)
What stands out is the -S(H2) you have. Shouldn't that be a +. If this doesn't fix the problem please repost this (at this place is more convenient) and copy the numbers you have in your table. You're going about this the right way. It's just a matter of getting the right numbers and right signs in the right place and punching in the right numbers on the calculator.

To find the entropy change (ΔS) for the given reaction, you need to use the standard molar entropy values for each component involved.

The correct formula to find ΔS of a reaction is:
ΔSrxn = ΣnS(products) - ΣmS(reactants)

Let's calculate the entropy change step by step:

1. Determine the stoichiometric coefficients for each substance in the balanced equation:
2 Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

2. Look up the standard molar entropies (S) for each component involved. Here are the values:
S(Na) = 51.21 J/(mol·K)
S(H2O) = 69.95 J/(mol·K)
S(NaOH) = 48.1 J/(mol·K)
S(H2) = 130.7 J/(mol·K)

3. Apply the formula:
ΔSrxn = (2×S(NaOH)) + (1×S(H2)) - (2×S(Na)) - (2×S(H2O))

Now, let's plug in the values and evaluate:
ΔSrxn = (2×48.1) + (1×130.7) - (2×51.21) - (2×69.95)
= 96.2 + 130.7 - 102.42 - 139.9
= 95.8 - 242.32
= -146.52 J/(mol·K)

Therefore, the correct entropy change for the given reaction is -146.52 J/(mol·K).

In your calculation, there may have been errors in substituting the values or mislabeling the signs of the entropies. Make sure to double-check the signs and values of the standard molar entropies used.