Given the reaction 2 Na2 02 (s) + 2 H2o(i) —- 4 NaOH(s) + 02(g) triangle H= -109 KJ

Question

determine the following reactions.
16 Na2o2(s)+ 16 H2o(i)———- 32 NaOH(s) + 8 O2(g)
2 NaOH(s)—-+ 1/2 o2(g)——-Nao2(s)+ H2o(I)

Answer 1

You really need to be more careful with your typing. I have an excuse when I make typos and omit characters; I'm old and can't see very well. You don't have that excuse. You're in a hurry. You can't hurry your way through chemistry. :-).
Equation 1 has dH = -109 kJ.
Look at equation 2. That is just 8x the first one; therefore dH must be 8*-109 kJ.

Look at equation 3. That one is 1/2 of the reverse of equation 1; therefore, dH is (1/2*-109 kJ) then change the sign since the equation is reversed. Note you made a typo here. I think you meant Na2O2(s)

Answer 2

My apologies. I will type it over.

Answer 3

Given the reaction 2 Na2 02 (s) + 2 H2o(i) —- 4 NaOH(s) + 02(g)

I couldn't make out the symbol however its triangle H= -109 KJ

determine the following reactions.
1. 16 Na2o2(s)+ 16 H2o(i)———- 32 NaOH(s) + 8 O2(g)
2. 2 NaOH(s)———— + 1/2 o2(g)——-Nao2(s)+ H2o(I)

Answer 4

The triangle is the Greek for delta.

Answer 5

Ok. I retyped the question straight from the book.hopefully its correct now.

Answer 6

Thanks. My answer stays the same. But I still think that last NaO2 is Na2O2. NaO2 doesn't make sense with the rest of the problem.