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 gas produced at 25 C and 1.07atm by the complete decomposition of 11.86 g of ammonium carbonate.
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2.91L
To calculate the total volume of gas produced by the complete decomposition of ammonium carbonate, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25°C + 273 = 298 K)
First, we need to determine the number of moles of ammonium carbonate (NH4)2CO3. We can do this by using the molar mass of (NH4)2CO3:
(NH4)2CO3:
NH4 = (1 x 1) + (4 x 1) = 18 g/mol
CO3 = (12 + (3 x 16)) = 60 g/mol
Total molar mass = (2 x 18) + 60 = 96 g/mol
Next, we calculate the number of moles of (NH4)2CO3:
moles = mass/molar mass
moles = 11.86 g / 96 g/mol = 0.1235 mol
According to the balanced equation, the decomposition of 1 mole of (NH4)2CO3 produces 2 moles of NH3 gas. Therefore, the number of moles of NH3 gas produced is twice the moles of (NH4)2CO3:
moles of NH3 = 2 x moles of (NH4)2CO3
moles of NH3 = 2 x 0.1235 mol = 0.247 mol
Now, we can use the Ideal Gas Law to calculate the volume of NH3 gas:
PV = nRT
V = (nRT) / P
V = (0.247 mol x 0.0821 L·atm/(mol·K) x 298 K) / 1.07 atm
Calculating the volume:
V = (6.47 L·atm) / 1.07 atm
V = 6.05 L
Therefore, the total volume of gas produced by the complete decomposition of 11.86 g of ammonium carbonate at 25°C and 1.07 atm is 6.05 liters.