Ammonium carbonate decomposes upon heating according to the following balanced equation:
(NH4)2CO3(s)→2NH3(g)+CO2(g)+H2O(g)
Calculate the total volume of gas produced at 24.0∘C and 1.02atm by the complete decomposition of 11.9g of ammonium carbonate.
Please show steps on how you got the answer. Thank you!
11.9g of am carb = 11.9/96.1 = 0.124 moles
So, you get
0.248 moles of NH3
0.124 moles CO2
0.124 moles H2O
Total moles of gas: 0.496
So, since 1 mole makes 22.4L at STP, adjust for your conditions.
If you have trouble adjusting for the conditions you can use PV = nRT. Plug in n from Steve's work, along with the other conditions, and solve for V in L. Same answer either way but I think this way is a little easier for students.
To calculate the total volume of gas produced by the complete decomposition of ammonium carbonate, we need to follow these steps:
Step 1: Calculate the moles of ammonium carbonate.
Given that the molar mass of (NH4)2CO3 is 96.09 g/mol, we can calculate the moles of ammonium carbonate using the formula:
Moles = Mass / Molar mass
Moles = 11.9 g / 96.09 g/mol
Moles = 0.124 mol
Step 2: Use the balanced equation to find the ratio of moles of gas produced.
According to the balanced equation, 1 mole of (NH4)2CO3 produces 2 moles of NH3, 1 mole of CO2, and 1 mole of H2O. Therefore, we have the following molar ratios:
NH3 : (NH4)2CO3 = 2 : 1
CO2 : (NH4)2CO3 = 1 : 1
H2O : (NH4)2CO3 = 1 : 1
Step 3: Calculate the moles of gas produced.
Using the molar ratios from step 2, we can calculate the moles of gas produced:
Moles of NH3 = 0.124 mol × 2 = 0.248 mol
Moles of CO2 = 0.124 mol × 1 = 0.124 mol
Moles of H2O = 0.124 mol × 1 = 0.124 mol
Step 4: Use the ideal gas law to calculate the volume of gas at 24.0∘C and 1.02 atm.
Given that the ideal gas law is written as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.
Rearranging the ideal gas law to solve for volume (V):
V = nRT / P
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 24.0 + 273.15
T(K) = 297.15 K
Using the given pressure of 1.02 atm and the gas constant R = 0.0821 L·atm/mol·K, we can now calculate the volume for each gas:
Volume of NH3 = (0.248 mol)(0.0821 L·atm/mol·K)(297.15 K) / (1.02 atm)
Volume of NH3 ≈ 6.68 L
Volume of CO2 = (0.124 mol)(0.0821 L·atm/mol·K)(297.15 K) / (1.02 atm)
Volume of CO2 ≈ 3.34 L
Volume of H2O = (0.124 mol)(0.0821 L·atm/mol·K)(297.15 K) / (1.02 atm)
Volume of H2O ≈ 3.34 L
Step 5: Calculate the total volume of gas produced.
The total volume of gas produced is the sum of the volumes of NH3, CO2, and H2O:
Total volume = Volume of NH3 + Volume of CO2 + Volume of H2O
Total volume = 6.68 L + 3.34 L + 3.34 L
Total volume = 13.36 L
Therefore, the total volume of gas produced at 24.0∘C and 1.02 atm by the complete decomposition of 11.9g of ammonium carbonate is approximately 13.36 L.