Using the balanced chemical equations obtained in the previous question, estimate the Q value for each explosive on a pe

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Using the balanced chemical equations obtained in the previous
question, estimate the Q value for each explosive on a per gram
basis. Use the following heats of formation (ΔHf) values
in kJ/mol.
(the previous questions)
Use the Springall Roberts rules to predict the products of the
explosive decomposition of the following (assume no additional
oxygen is available):
Step 1:
C5H8N4O12 - 5CO
7 Oxygen are left
Step 2:
C5H8N4O12 - 5CO +4H2O
All hydrogens are used.
3 oxygen will be left
Step 3:
C5H8N4O12 - 3CO2 + 2CO +4H2O
All oxygens are used.
All hydrogens are used.
Step 4:
C5H8N4O12 - 3CO2 + 2CO +4H2O +2 N2
FINALLY - C5H8N4O12 - 3CO2 + 2CO +4H2O +2 N2
So, number of Carbon, C= 3+2= 5
So, number of Hydrogen = 4 x 2 = 8.
So, number of Nitrogen, N = 2 x 2 = 4
So, number of Oxygen, O = 3 x 2 +2 +4 = 12.
Balanced by the SR rules, with no additional oxygen
provided.
Step 1:
C2H4N2O6 - 2CO
4 oxygen are left.
Step 2:
C2H4N2O6 - 2CO + 2H2O
All hydrogens are used.
2 oxygen will be left.
Step 3:
C2H4N2O6 - 2CO + 2H2O
All Hydrogens and Oxygen are used.
Step 4:
C2H4N2O6 - 2CO2 + N2 + 2H2O
All Hydrogens, Nitrogen and Oxygens are used.
FINALLY - C2H4N2O6 - 2CO2 + N2 + 2H2O
So, number of Carbon, C = 2.
So, number of Hydrogen, H = 2 x 2 =4
So, number of Nitrogen, N = 2
So, number of Oxygen O = 2 x 2 + 2 = 6
Balanced by the SR rules, with no additional oxygen
provided.