molar mass C3H5N3O9 = 227
dH = 6.26 kJ/g x (227 g/mol) = 1421 kJ/mol
4C3H5(NO3)3 → 12CO2 + 6N2 + 10H2O + O2 dH = -1421 kJ/mol
29 mol products are formed for every 4 mol nitroglycerin detonated.
So 4.65 mol products must have come from 4.65 x (4/29) = 0.138 mol nitroglycerin.
1421 kJ/mol x 0.138 mol = ? kJ = dH for 4.65 mols products.
Nitroglycerine, C3H5(NO3)3(l), is an explosive most often used in mine or quarry blasting. It is a powerful explosive because four gases (N2, O2, CO2, and steam) are formed when nitroglycerine is detonated. In addition, 6.26 kJ of heat is given off per gram of nitroglycerine detonated. (a) Write a balanced thermochemical equation for the reaction. (b) What is ΔH when 4.65 mol of products is formed?
2 answers
oops
Forgot to add the states. The thermochemical equation will be
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) + 5684 kJ OR
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) dH = -1421 kJ/mol OR
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) dH = -5684 kJ
Forgot to add the states. The thermochemical equation will be
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) + 5684 kJ OR
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) dH = -1421 kJ/mol OR
4C3H5(NO3)3(l) → 12CO2(g) + 6N2(g) + 10H2O(g) + O2(g) dH = -5684 kJ