Ammonium carbonate decomposes upon heating according to the following balanced equation: (NH4)2CO3 (s) → 2 NH3 (g) + CO2 (g) + H2O (g) Calculate the total volume of ammonia gas produced at 25.5 oC and 1 atm by the complete decomposition of 15 g of ammonium carbonate.

7.63 L is the produced volume of ammonia.

Explanation:

The balanced reaction of decomposition is this:

(NH₄)₂CO₃ (s) →  2NH₃ (g) + CO₂ (g) + H₂O (g)

We convert the mass of carbonate to moles → 15 g . 1mol / 96g = 0.156 mol

Ratio is 1:2, so the 0.156 moles will produce the double of moles of ammonia. → 0.156 . 2 = 0.312 moles

We use now,  the Ideal Gases Law to determine the volume

Pressure . Volume = moles . R . T

T → 25.5°C + 273  = 298.5K

1 atm . V = 0.312 mol . 0.082 L.atm/mol.K . 298.5K

V = (0.312 mol . 0.082 L.atm/mol.K . 298.5K) / 1 atm → 7.63 L

The total volume of ammonia gas is 7.65 L

Explanation:

Step 1: Data given

Temperature = 25.5 °C

Pressure = 1.0 atm

Mass of ammonium carbonate = 15.0 grams

Molar mass ammonium carbonate = 96.09 g/mol

Step 2: The balanced equation

(NH4)2CO3 (s) → 2 NH3 (g) + CO2 (g) + H2O (g)

Step 3: Calculate moles of (NH4)2CO3

Moles (NH4)2CO3 = mass (NH4)2CO3 / molar mass (NH4)2CO3

Moles (NH4)2CO3 = 15.0 grams / 96.09 g/mol

Moles (NH4)2CO3 = 0.156 moles

Step 4: Calculate moles NH3

For 1 mol (NH4)2CO3 we'll have 2 moles NH3, 1 mol CO2 and 1 mol H2O

For 0.156 moles (NH4)2CO3 we'll have 2*0.156 = 0.312 moles NH3

Step 5: Calculate volume of NH3 gas

p*V = n*R*T

⇒ with p = the pressure = 1 atm

⇒ with V = the volume of NH3 gas = TO BE DETERMINED

⇒ with n = the moles of NH3 gas = 0.312 moles

⇒ with R = the gas constant = 0.08206 L*atm/ mol*K

⇒ with T = The temperature = 25.5 °C = 298.65 K

V = (n*R*T)/p

V = (0.312 * 0.08206 * 298.65) / 1

V = 7.65 L

The total volume of ammonia gas is 7.65 L