*If the density of TNT is 1.65 g/cm^3, calculate the volume of a 454 g (which is

Question

*If the density of TNT is 1.65 g/cm^3, calculate the volume of a 454 g (which is 1 pound) quantity of the explosive before the explosion and the volume of gas produced after the explosion. Assume that the volume of the C(s) in the product of the following reaction is negligible (meaning you only need to focus on the 15 mol of gaseous products), and assume STP conditions: 2C7H5N3O6 (l) --> 7CO (g) + 7C (s) + 5H2O (g) + 3N2 (g) All numbers immediately following elemental symbols are subscripts. You must show all work to receive credit. (HINT: You can use density, and the fact that 1 cm^3 = 1 mL, to find the initial condition for volume of the liquid. You can then use the mol of gaseous products to determine volume of final condition since you know molar volume at STP.)

Explanation / Answer

Balanced Chemical Equation:

2C7H5N3O6(l) = 7CO(g) + 7C(s) + 5H2O(g) + 3N2(g)

Volume of TNT before explosion:

454 g x 1 cm^3/1.65 g x 1 mL/1 cm^3 = 275 mL

Volume of TNT after explosion:

First, solve for the moles of TNT:

454 g x 1 mol/227.13 g = 2.00 moles

Next, solve for the moles of gaseous products from the balanced chemical equation:

2.00 moles of TNT x 7 moles of CO/2 moles of TNT = 7 moles of CO

2.00 moles of TNT x 5 moles of H2O/2 moles of TNT = 5 moles of H2O

2.00 moles of TNT x 3 moles of N2/2 moles of TNT = 3 moles of N2

Add up the moles of gasesous products: 7 + 5 + 3 = 15 moles

Use PV = nRT to solve for the volume:

@STP, T = 273 K & P = 1 atm

n = 15 moles

R = 0.0821 atm*L/mol*K

V = ?

PV = nRT

V = nRT / P

V = (15 moles)(0.0821 atm*L/mol*K)(273 K) / 1 atm

V = 336 L or 0.336 mL

Hope this helps! :)

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