When heated to 350 °C at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases: 2NH...

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When heated to 350 °C at 0.950 atm, ammonium nitrate decomposes to produce nitrogen, water, and oxygen gases: 2NH, NO3(s) —2N

Part A How many liters of water vapor are produced when 22.6 g of NH, NO, decomposes? Express your answer with the appropriat

X Incorrect; Try Again; 4 attempts remaining Part B How many grams of NH, NO3 are needed to produce 13.0 L of oxygen? Express

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Answer 1

Answer #1

A)

Molar mass of NH4NO3,

MM = 2*MM(N) + 4*MM(H) + 3*MM(O)

= 2*14.01 + 4*1.008 + 3*16.0

= 80.052 g/mol

mass of NH4NO3 = 22.6 g

mol of NH4NO3 = (mass)/(molar mass)

= 22.6/80.05

= 0.2823 mol

According to balanced equation

mol of H2O formed = (4/2)* moles of NH4NO3

= (4/2)*0.2823

= 0.5646 mol

Given:

P = 0.95 atm

n = 0.5646 mol

T = 350.0 oC

= (350.0+273) K

= 623 K

use:

P * V = n*R*T

0.95 atm * V = 0.5646 mol* 0.08206 atm.L/mol.K * 623 K

V = 30.3834 L

Answer: 30.4 L

B)

Given:

P = 0.95 atm

V = 13.0 L

T = 350.0 oC

= (350.0+273) K

= 623 K

find number of moles using:

P * V = n*R*T

0.95 atm * 13 L = n * 0.08206 atm.L/mol.K * 623 K

n = 0.2416 mol

From reaction,

Mol of NH4NO3 reacted = 2*mol of O2 produced

= 2 * 0.2416 mol

= 0.4832 mol

Molar mass of NH4NO3,

MM = 2*MM(N) + 4*MM(H) + 3*MM(O)

= 2*14.01 + 4*1.008 + 3*16.0

= 80.052 g/mol

use:

mass of NH4NO3,

m = number of mol * molar mass

= 0.4832 mol * 80.05 g/mol

= 38.68 g

Answer: 38.7 g

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Answer 2

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