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.
See earlier response.
To calculate the total volume of gas produced by the decomposition of ammonium carbonate, we need to follow several steps:
Step 1: Convert the mass of ammonium carbonate to moles.
To find the number of moles, we divide the given mass (11.86 g) by the molar mass of ammonium carbonate. The molar mass can be calculated by adding up the atomic masses of each element in the compound.
(NH4)2CO3:
(2 * 14.01 g/mol) + (4 * 1.01 g/mol) + 12.01 g/mol + (3 * 16.00 g/mol) = 96.09 g/mol
Number of moles = 11.86 g / 96.09 g/mol ≈ 0.1234 mol
Step 2: Use the mole ratio to determine the moles of gas produced.
From the balanced equation, we can see that for every 1 mole of (NH4)2CO3, 2 moles of NH3, CO2, and H2O are produced.
Therefore, the moles of gas produced = 0.1234 mol * 2 = 0.2468 mol
Step 3: Apply the ideal gas law to calculate the volume of gas.
The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
Pressure (P) = 1.07 atm
Temperature (T) = 25 °C = 298 K (converted to Kelvin)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
Rearranging the ideal gas law equation:
V = (n * R * T) / P
= (0.2468 mol * 0.0821 L·atm/(mol·K) * 298 K) / 1.07 atm
≈ 5.62 L
Hence, 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 approximately 5.62 liters.