The combustion of TNT (trinitrotoluene. MM=227.13) in oxygen is given by the fol

Question

The combustion of TNT (trinitrotoluene. MM=227.13) in oxygen is given by the following equation: C_7H_5N_3O_6(s) + O_2(g) rightarrow N_2(g) + H_2O (g) + CO_2 (g) (not balanced). Calculate the total, volume (in L) of gas (at 152.0 degree C and 135.5 kPa) produced by the complete decomposition of 1.850 kg of trinitrotoluene, (assume reaction carried out in excess O_2) L Tries 0/2 What Is the partial pressure of H_2O and CO_2 (assume all O_2 are completely reacted? P_H_2O atm Tries 0/2 P_CO_2 atm Tries 0/2

Explanation / Answer

4C7H5N3O6+ 21O2 28CO2+6N2+10H2O (Balanced equation)

Amount of TNT = 1.850 kg = 1850g

Molecular weight to TNT: 227.13 g/mol

no. of moles of TNT = 1850/227.13 = 8.145 moles

4 moles of TNT gives 44 moles of gaseous products (6N2, 28CO2 and 10H2O)

8.145 moles of TNP will produce 89.5 moles of gaseous products.

P = 135.5 kPa (1kPa = 0.00986 and hence 135.5kPa = 1.33 atm)

T = 152 + 273 = 425K

V =?

R = 0.0821LatmK-1mol-1

using ideal gas equation: PV = nRT

V = nRT/P = 89.5 moles x 0.0821LatmK-1mol-1 x 425K /1.33 atm = 2348L

Volume of the gas producted by combustion is 2348L

2. Partial pressure of H2O and CO2.

PH2O = xH2O * Ptotal

Ptotal = 1.33 atm

total number of moles of gaseous products: 44

No. of moles of H2O produced: 10 moles

Mole fraction of water vapor = 10/44

XH2O = (10/44) * 1.33 = 0.3 atm

No. of moles of CO2 produced: 28 moles

Mole fraction of CO2 = 28/44

PH2O = xH2O * Ptotal

Ptotal = 1.33 atm

XCO2 = (28/44) * 1.33 = 0.85 atm

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