What volume of gas can be produced in a lab where pressure is 16.0 psi and temperature is 25 C if you collect it over water from the reaction of 50g of ammonium sulfate with 50g of aluminum hydroxide. The equation for the reaction is:
3(NH4)2SO4 + 2Al(OH)3 ---> Al2(SO4)3 + 6NH3 + 6H2O
To calculate the volume of gas produced in this reaction, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (Note: 1 atm = 14.7 psi)
V = volume in liters
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K) or 8.314 J/(mol·K))
T = temperature in Kelvin (K)
First, let's calculate the number of moles of ammonia (NH3) produced in the reaction:
From the balanced equation:
3(NH4)2SO4 + 2Al(OH)3 ---> Al2(SO4)3 + 6NH3 + 6H2O
We see that for every 3 moles of ammonium sulfate (NH4)2SO4, we produce 6 moles of ammonia (NH3). So, the number of moles of ammonia produced can be calculated as follows:
50 g of ammonium sulfate (NH4)2SO4 * (1 mol / 132.14 g) * (6 mol NH3 / 3 mol (NH4)2SO4) = (50/132.14) * (6/3) mol
In the same way, we can calculate the number of moles of ammonia produced:
50 g of aluminum hydroxide Al(OH)3 * (1 mol / 78.0 g) * (6 mol NH3 / 2 mol Al(OH)3) = (50/78) * (6/2) mol
Now, we can calculate the total number of moles of ammonia produced by adding the moles from the reactants:
Total moles of NH3 produced = [(50/132.14) * (6/3)] + [(50/78) * (6/2)] mol
Next, we need to convert the temperature to Kelvin. The temperature given is 25 degrees Celsius, which can be converted as follows:
T(K) = 25°C + 273.15 = 298.15 K
Now, we can calculate the volume of gas produced using the ideal gas law:
PV = nRT
V = (nRT) / P
V = [(total moles of NH3) * (gas constant) * (temperature)] / pressure
V = [(total moles of NH3) * (0.0821 L·atm/(mol·K)) * (298.15 K)] / (16.0 psi * (1 atm / 14.7 psi))
Now, plug in the values and calculate:
V = [(total moles of NH3) * (0.0821 L·atm/(mol·K)) * (298.15 K)] / (16.0 psi * (1 atm / 14.7 psi))
Note: Make sure to use consistent units throughout the calculation.
Finally, substitute the previous value to find the volume of gas produced.
To find the volume of gas produced, we need to use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas (in this case, 16.0 psi)
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature of the gas (in this case, 25°C = 298 K)
First, we need to calculate the number of moles of the gas produced. We can do this by finding the limiting reactant in the reaction.
1. Calculate the number of moles of ammonium sulfate (NH4)2SO4:
molar mass of (NH4)2SO4 = 132.14 g/mol
moles of (NH4)2SO4 = mass / molar mass = 50 g / 132.14 g/mol
2. Calculate the number of moles of aluminum hydroxide Al(OH)3:
molar mass of Al(OH)3 = 78.0 g/mol
moles of Al(OH)3 = mass / molar mass = 50 g / 78.0 g/mol
3. To determine the limiting reactant, compare the moles of each reactant to their stoichiometric coefficients in the balanced equation:
From the balanced equation: 3(NH4)2SO4 + 2Al(OH)3 ---> Al2(SO4)3 + 6NH3 + 6H2O
Ratio of moles: (NH4)2SO4 : Al(OH)3 = 3:2
Divide the moles of each reactant by their stoichiometric coefficients to get the number of moles per coefficient:
(NH4)2SO4: 50 g / 132.14 g/mol = 0.378 mol
Al(OH)3: 50 g / 78.0 g/mol = 0.641 mol
The mole ratio is 0.378 mol : 0.321 mol ≈ 1.18 : 1
Since Al(OH)3 has fewer moles for the same amount of mass, it is the limiting reactant.
4. Calculate the moles of ammonia (NH3) produced:
From the balanced equation: 2Al(OH)3 ---> 6NH3
The stoichiometric ratio is 2:6.
moles of NH3 = 6 × moles of Al(OH)3 = 6 × 0.641 mol
5. Now, we can plug the values into the ideal gas law equation:
PV = nRT
V = nRT / P
V = (6 × 0.641 mol) × (0.0821 L·atm/mol·K) × (298 K) / 16.0 psi
After performing the calculation, you will find the volume of gas produced in the lab.