P4(s) + 3o2(g) → p4o6(s) δh1 = -1,640.1 kj p4o10(s) → p4(s) + 5o2(g) δh2 = 2,940.1 kj what is the overall enthalpy of reaction for the equation shown below? round the answer to the nearest whole number. p4o6(s) + 2o2(g) -> p4o10(s)

P4(s) + 3O2(g) → P4O6(s)     ΔH1 = -1,640.1 kJ
P4O10(s) → P4(s) + 5O2(g)     ΔH2 = 2,940.1 kJ 
P4O6(s) + 2O2(g) -> P4O10(s)     ΔH=?

Solution:
P4O6(s) → P4(s) + 3O2(g)     ΔH = 1,640.1 kJ
P4(s) + 5O2(g) → P4O10(s)     ΔH = -2,940.1 kJ 
+
----------------------------
P4O6(s) + 5O2(g) -> P4O10(s) ΔH = -1,3 kJ

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.