What is the increase in entropy when 11.2 l of o2 are mixed with 11.2 l of h2 at stp?

Answer 1

Is the water gas or liquid. I wrote (g) but you should change it if necessary ans use the appropriate numbers.
2H2(g) + O2(g) => 2H2O(g)
What is the limiting reagent (LR)?
11.2 L H2 will require 5,6 L O2 and you have that much O2 so H2 is the LR.
Start ............11.2 L H2 + 11.2 L = 0 L H2O
change........-11.2..............-5.6 L.....11.2 L
equil................0..................5.6 L.....11.2 L
dSrxn = (n*So products) - (n*So reactants)
Post your work if you get stuck.