Question

What mass of potassium nitrate is needed to generate 157.0 L of gas, composed of 105.0 L of N2 and 52.0 L of O2 at 0.920 atm and 293 K, using these two reactions?

Answer 1

Nitrogen can be produced from potassium nitrate as indicated by the balanced equation:

10Na + 2KNO₃ → K₂O + 5Na₂O + N₂

PV = nRT

V(N₂) = 105 L

P = 0.920 atm

T = 293 K

R = 0.08206 L×atm/mol×K

$n = \frac{PV}{RT}$

n(N₂) $= \frac{0.920 \times 105}{0.08206 \times 293} = 4.02 \;mol$

According to the equation:

n(KNO₃) = 2n(N₂) $= 2 \times 4.02 = 8.04 \;mol$

M(KNO₃) = 101.1 g/mol

$m = n\times M$

m(KNO₃) $= 8.04 \times 101.1 = 812.84 \;g$

Oxygen gas can be generated by heating potassium chlorate as indicated by the balanced equation:

2KClO₃ → 2KCl + 3O₂

V(O₂) = 52.0 L

n(O₂) $= \frac{0.920 \times 52}{0.08206 \times 293} = 1.99 \;mol$

According to the equation:

n(KClO₃) = 2n(O₂) $= 2 \times 1.99 = 3.98 \;mol$

M(KClO₃) = 122.55 g/mol

m(KClO₃) $= 3.98 \times 122.55 = 487.75 \;g$