Question

When heated at 405 C at 0.850 atm, ammonium nitrate decomposes to produce nitrogen, water and oxygen gases. using the ideal gas law equation, calculate the volume in milliliters, of water vapor produced when 85.0 g of NH4NO3 decomposes?

2 NH4NO3 (s) ---> 2N2 (g) +4 H2O (g) + O2 (g)

Answer 1

Answer on Question #49592 - Chemistry - Other

Question

When heated at 405 C at 0.850 atm, ammonium nitrate decomposes to produce nitrogen, water and oxygen gases. using the ideal gas law equation, calculate the volume in milliliters, of water vapor produced when 85.0 g of NH₄NO₃ decomposes?

$$2 \mathrm{NH_4NO_3}(s) \rightarrow 2 \mathrm{N_2}(g) + 4 \mathrm{H_2O}(g) + \mathrm{O_2}(g)$$

Answer:

Number of moles of NH₄NO₃ is:

$$\frac{m}{M} \quad \frac{85.0}{80} \quad 1.06 \text{ moles}$$

According to the reaction, 2 moles of NH₄NO₃ produce 4 moles of H₂O, therefore, 1.06 moles of NH₄NO₃ produce 2.12 moles of H₂O.

Ideal gas law equation is:

$$PV \quad RT$$

Then the volume of water vapor produced when 85.0 g of NH₄NO₃ decomposes is:

$$V \quad \frac{RT}{P} \quad \frac{2.12 \cdot 0.082 \cdot (405 + 273)}{0.850} \quad 138.663 \text{ L} \quad 138663 \text{ mL}$$

Answer: 138663 mL

http://www.AssignmentExpert.com/