Question

You are given the following data.
P4(s) + 6 Cl2(g) 4 PCl3(g) H = -1225.6 kJ
P4(s) + 5 O2(g) P4O10(s) H = -2967.3 kJ
PCl3(g) + Cl2(g) PCl5(g) H = -84.2 kJ
PCl3(g) + 1/2 O2(g) Cl3PO(g) H = -285.7 kJ

Calculate H for the following reaction.
P4O10(s) + 6 PCl5(g) 10 Cl3PO(g)

Answer 1

Do you omit the arrows on purpose? I can't tell reactants from products without the arrows. Please repost and show the arrows.

Answer 2

You are given the following data.
P4(s) + 6 Cl2(g)--> 4 PCl3(g) H = -1225.6 kJ
P4(s) + 5 O2(g) --> P4O10(s) H = -2967.3 kJ
PCl3(g) + Cl2(g) --> PCl5(g) H = -84.2 kJ
PCl3(g) + 1/2 O2(g) --> Cl3PO(g) H = -285.7 kJ

Calculate H for the following reaction.
P4O10(s) + 6 PCl5(g) --> 10 Cl3PO(g)

i got -610.1kJ is that correct?

Answer 3

I obtained the same number of 610.1 kJ.

Answer 4

and it was a positive, right?

Answer 5

No. I think it is a negative 610.1 kJ.
Equations 2 and 3 are reversed. Equations 1 and 4 are as written. Equation 3 is multiplied by 6 and equation 4 is multiplied by 10. Check my numbers.

Answer 6

yep you are right thanks