Question

Igniting gunpowder produces nitrogen and carbon dioxide gas that propels a bullet by the reaction

2KNO3 (s) + 1/8S8(s) +3C(s) --> K2S(s) +N2(g) + 3 CO2(g)

If 3.15 g of KNO3 reacts with sufficient sulfur (S8) and carbon (C), how much P-V work will the gases do against an external pressure of 1.00 atm given the densities of nitrogen and carbon dioxide are 1.165 g/L and 1.830 g/L, respectively, at 20°C?

Answer 1

mass of KNO3 = 3.15 grams

molar mass of KNO3 = 101.1 g/mol so

moles of KNO3 = mass / molar mass = 3.15 / 101.1 = 0.03116 moles of KNO3

Acording to the reaction

2 mole of KNO3 will form 1 mole of N2 so

0.03116 moles of KNO3 = 0.0155 moles of N2

molar mass of N2 is 28 g/gmol

mass of N2 is 0.0155 * 28 = 0.436 grams

calculate volume of N2

volume = mass / density = 0.436 / 1.165 = 0.3744 liters of n2 formed

For CO2, reaction is 1 : 3

0.03116 moles of KNO3 will produce 3/2 * 0.03116 = 0.0467 moles of CO2

calculate mass of co2, molar mass of CO2 is 44

mass of CO2 = 44 * 0.04673 = 2.05638 grams of CO2 formed

calculate volume

mass / density = 2.05638 / 1.83 = 1.123 liters of CO2 formed

Total volume of gas formed

0.374+1.123 = 1.4981 liters

initial volume of gas is 0

turn liters to m3, do this dividing by 1000 so

volume = 0.0014981 m3

turn atm to pascals

1 atm = 101 325 Pa

w = P (Vf-Vi) = 101 325 * (0.0014981 - 0) = 151.79 Joules

where w is work

p is pressure

V is volume

f and i are for final and initial respectively

remember that Pascals * m³ = Joules

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